Question

An AND gate is followed by a NOT gate in series. With two inputs ' \(A\) ' and ' \(B\) ', the Boolean expression for the output ' \(Y\) ' will be

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

Hint:

An AND gate followed by a NOT gate is equivalent to a single NAND gate.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question describes a compound logic gate. We need to find the final logic expression by combining the operations of individual gates.
Step 3: Detailed Explanation:
1) The first gate is an AND gate. Its output, say \(X\), for inputs \(A\) and \(B\) is given by:
\[ X = A \cdot B \]
2) This output \(X\) is then passed through a NOT gate. The final output \(Y\) is the inversion of its input:
\[ Y = \overline{X} \]
3) Substituting \(X\):
\[ Y = \overline{A \cdot B} \]
This combination is fundamentally a NAND gate.
Step 4: Final Answer:
The Boolean expression is \(\overline{A \cdot B}\).