The truth table for the following logic circuit is:

Solution & Explanation

1. Trace the Inputs to the AND Gates

Let's track the binary signals moving from the input terminals through the branches:

• Top AND Gate:
  • The upper input branch connects directly to the line coming from terminal $$A$$.
  • The lower input branch passes through an inverter (NOT gate) connected crosswise from terminal $$B$$, which yields $$\bar{B}$$.
  • Thus, the expression for the upper gate's output is: $$\text{Out}_{\text{top}} = A \cdot \bar{B}$$
• Bottom AND Gate:
  • The upper input branch passes through an inverter (NOT gate) connected crosswise from terminal $$A$$, which yields $$\bar{A}$$.
  • The lower input branch connects directly to the line coming from terminal $$B$$.
  • Thus, the expression for the lower gate's output is: $$\text{Out}_{\text{bottom}} = \bar{A} \cdot B$$

2. Evaluate the Final NOR Gate

Both of these intermediate outputs flow directly into a two-input NOR gate at the final stage. A NOR gate adds its inputs together (OR operation) and then inverts the entire sum:

$$Y = \overline{\text{Out}_{\text{top}} + \text{Out}_{\text{bottom}}}$$

$$Y = \overline{(A \cdot \bar{B}) + (\bar{A} \cdot B)}$$

Notice that the inner term $$(A \cdot \bar{B}) + (\bar{A} \cdot B)$$ represents a standard XOR gate function ($$A \oplus B$$). Since it is inverted by the outer overbar, the complete expression is equivalent to an XNOR gate ($$\overline{A \oplus B}$$):

$$Y = A \odot B$$

3. Construct the Truth Table

An XNOR gate outputs a high logical state ($$1$$) only when both inputs match each other (i.e., both are $$0$$ or both are $$1$$):

Correct Option: The truth table option displaying output states $$[1, 0, 0, 1]$$ sequentially from top to bottom is correct ..