Question

In the combination of the following gates the output $Y$ can be written in terms of inputs $A$ and $B$ as

• A. $\overline{A.B}$
• B. $A . \overline{B} + \overline{A} . B$
• C. $\overline{A.B} + A .B $
• D. $\overline{A + B}$

Answer 1

The Correct Option is B

To determine the output $Y$ in terms of inputs $A$ and $B$, we need to analyze the given options and identify which Boolean expression corresponds to common logic gates and their combinations.

The provided options are:

1. $\overline{A.B}$ - This is the expression for a NAND gate, which outputs the logical negation of the AND operation between $A$ and $B$.
2. $A . \overline{B} + \overline{A} . B$ - This is the expression for an XOR gate. The XOR gate outputs true only if the number of true inputs is odd. Specifically, it is true if one of the inputs is true and the other is false.
3. $\overline{A.B} + A .B $ - This combination represents a logical tautology, which is always true, regardless of the input values.
4. $\overline{A + B}$ - This is the expression for a NOR gate, providing the negation of the OR operation between $A$ and $B$.

Given the correct answer is $A . \overline{B} + \overline{A} . B$, let us understand why this is valid:

Explanation:

• An XOR gate outputs true if exactly one of its inputs is true. The expression $A . \overline{B} + \overline{A} . B$ satisfies this condition:
• When $A = 1$ and $B = 0$, the output is $1$ because $A . \overline{B}$ evaluates to $1$.
• When $A = 0$ and $B = 1$, the output is $1$ because $\overline{A} . B$ evaluates to $1$.
• If both inputs are the same (either $0,0$ or $1,1$), neither term in the expression is true, resulting in an output of $0$.

Therefore, the correct output expression for the XOR configuration of logic gates with inputs $A$ and $B$ is $A . \overline{B} + \overline{A} . B$.

This understanding solidifies why the correct answer is indeed the XOR representation, and it is consistent with standard digital electronics theory.