Question

What is the Boolean equation for a NAND gate with inputs \(A\) and \(B\)?}

• A. \(Y = A + B\)
• B. \(Y = \overline{A+B}\)
• C. \(Y = \overline{A} + \overline{B}\)
• D. \(Y = \overline{A \cdot B}\)

Hint:

De Morgan's Theorem: \(\overline{AB} = \overline{A} + \overline{B}\) and \(\overline{A+B} = \overline{A}\cdot\overline{B}\). NAND is the complement of AND.

Answer 1

The Correct Option is D

Step 1: Basic Principle
A NAND gate produces the complement of an AND operation.
Step 2: Solution Procedure:
By definition: NAND gate output is \(\overline{A \cdot B}\). By De Morgan's theorem, this is equivalent to \(\overline{A} + \overline{B}\), but the Boolean equation is \(\overline{A \cdot B}\).
Step 3: Required Answer:
Boolean equation: \(Y = \overline{A \cdot B}\).