7485 magnitude comparator — determine outputs from given inputs and cascade (expansion) signals Given a 7485 4-bit magnitude comparator with the following values: A0 = 0, B0 = 1, (IA < B) = 0; A1 = 1, B1 = 0, (IA = B) = 1; A2 = 1, B2 = 0, (IA > B) = 0; A3 = 0, B3 = 0. What are the states of the three outputs A = B, A < B, and A > B?

Introduction / Context:The 7485 is a 4-bit magnitude comparator providing three primary outputs: A > B, A = B, and A < B. It also includes cascade (expansion) inputs to chain devices for wider word lengths. Correctly interpreting bit significance and the expansion inputs is essential to determine the final comparison result.

Given Data / Assumptions:

• A3..A0 and B3..B0 represent 4-bit words with A3/B3 as the most significant bits.
• Expansion inputs are: (IA < B), (IA = B), (IA > B) = 0, 1, 0 respectively, meaning higher-order sections are equal.
• Bit values provided: A3 = 0, B3 = 0; A2 = 1, B2 = 0; A1 = 1, B1 = 0; A0 = 0, B0 = 1.

Concept / Approach:When the expansion input IA = B = 1, the present 4-bit section decides the result. Evaluate from MSB to LSB: the first bit position where A and B differ determines the outcome. If they are equal through all bits, output A = B = 1; otherwise assert A > B or A < B at the first differing position with A or B larger.

Step-by-Step Solution:

Verification / Alternative check:Interpreting A and B as unsigned: A = 0b0110x (given lower bits) is larger than B since A2 = 1 and B2 = 0 at the first differing position. This aligns with the standard priority of higher-order bits in magnitude comparison.

Why Other Options Are Wrong:

• Any option asserting A < B contradicts the MSB-first comparison (A2 > B2).
• Asserting A = B is impossible because multiple bits differ.
• All zeros or all ones outputs cannot occur for valid comparisons with IA = B = 1.

Common Pitfalls:

• Reading bits LSB-to-MSB; must evaluate MSB-first.
• Ignoring expansion inputs; IA = B must be 1 to let this stage decide.

Final Answer:A = B = 0, A < B = 0, A > B = 1