Complete state decoding for a MOD-64 counter — gate count and inputs How many AND gates are required to decode every state of a MOD-64 counter, and how many inputs must each AND gate have?

Introduction / Context:Decoding a counter state means generating a unique output that is HIGH for exactly one binary pattern and LOW otherwise. This is common in address decoding, time-slot generation, and sequence control. For an n-bit binary counter, each state can be decoded using an n-input AND (or NAND with polarity adjustments) that matches the required combination of bits and complements.

Given Data / Assumptions:

• MOD-64 counter ⇒ 64 states ⇒ requires 6 bits (since 2^6 = 64).
• Full decoding implies one line per state (one-hot outputs).
• Assume positive-logic AND decoding with complemented inputs where needed.

Concept / Approach:Each distinct 6-bit value (b5…b0) must be recognized. A single 6-input AND gate, fed by the appropriate direct or inverted bit lines, asserts HIGH only when its 6 inputs match the target pattern. Repeating this for all 64 patterns yields a full decoder.

Step-by-Step Solution:

Verification / Alternative check:Compare to standard parts: a 6-to-64 decoder (e.g., built from smaller 3-to-8 blocks) provides 64 one-hot outputs; internally, the logic effectively realizes 64 minterms over 6 variables—conceptually equivalent to 64 AND terms of 6 inputs each.

Why Other Options Are Wrong:

• 128 gates choices: would duplicate lines unnecessarily; only 64 unique states exist.
• 5 inputs per gate: cannot uniquely identify all 6-bit patterns.
• 32 gates: insufficient to cover 64 states.

Common Pitfalls:

• Forgetting to account for complemented inputs; practical designs often use NAND/NOR plus inverters.

Final Answer:64 gates, 6 inputs to each gate