In digital electronics, how can you detect whether two 8-bit binary signals are exactly the same on all bit positions?

Introduction / Context:
In digital electronics and computer hardware, it is very common to compare two binary words to decide whether they are exactly equal. For example, a CPU may compare an operand with a stored value, or a communication interface may check whether received data matches what was transmitted. For 8-bit signals, we need a simple and reliable way to detect equality across all eight bit positions. This question tests your understanding of basic comparator design using standard logic gates.

Given Data / Assumptions:

• There are two 8-bit binary signals that we can call A7..A0 and B7..B0.
• We want a single output that is high only when all corresponding bits are equal.
• Only basic combinational logic elements such as AND, OR, XOR, and XNOR gates are assumed.
• We want a straightforward hardware implementation that can be built from simple gates.

Concept / Approach:
To detect whether two individual bits are the same, we can use an XNOR gate. An XNOR outputs logic 1 when both inputs are equal, either 0 and 0 or 1 and 1. To compare two 8-bit words, we can place an XNOR on each pair of corresponding bits. If all eight XNOR outputs are 1, then the two words are identical. A multi-input AND gate can combine the eight XNOR outputs into a single equality signal that is 1 only if every bit matches.

Step-by-Step Solution:
Step 1: For each bit position i from 0 to 7, connect Ai and Bi to an XNOR gate so that each gate outputs 1 when the two bits at that position are equal.Step 2: Collect the eight XNOR outputs into a multi-input AND gate or a tree of smaller AND gates.Step 3: The output of the final AND gate will be high only when all individual XNOR outputs are high.Step 4: Therefore the AND output is a single equality flag that indicates whether the two 8-bit signals are exactly the same.Step 5: Recognize that this is the standard way to build a digital comparator for equality checking.

Verification / Alternative check:
Consider two simple test cases. If both 8-bit inputs are 10101010, then every XNOR gate sees identical bits and outputs 1. The AND gate then outputs 1, signalling equality. If one signal is 10101010 and the other is 10101110, at least one bit position is different. The XNOR at that position outputs 0, and the AND gate output becomes 0, signalling inequality. This behaviour exactly matches the requirement for an equality detector, so the proposed circuit is correct.

Why Other Options Are Wrong:
Option B is incorrect because OR gates in series do not detect equality; they simply propagate any high inputs and cannot distinguish matching from mismatching bit patterns. Option C is wrong because counting pulses with an up-counter does not compare static 8-bit values directly and is unnecessarily complex for this task. Option D is incorrect because a single D flip-flop cannot compare all eight bits simultaneously and is meant for storage or synchronization, not for multi-bit comparison.

Common Pitfalls:
Students sometimes try to use XOR gates directly and forget that XOR outputs 1 on inequality, not equality, so its output must be inverted or combined carefully. Another common mistake is to compare only one or two bits and assume that is enough for equality, which is not correct for full 8-bit comparison. It is also easy to confuse magnitude comparators, which tell you greater than or less than, with simple equality comparators. Remember that XNOR is the key gate for equality testing, and combining multiple XNOR outputs with an AND gate scales this idea to multi-bit signals.

Final Answer:
The correct hardware method is to use eight XNOR gates, one for each bit, feeding a multi-input AND gate so the output is high only when all bits match.