Difficulty: Easy
Correct Answer: NAND
Explanation:
Introduction / Context:The sum-of-products (SOP) form is one of the most common ways to describe Boolean functions for hardware realization. Engineers frequently translate an SOP expression into a two-level gate network. Choosing a universal gate family simplifies the build, reduces part types, and eases optimization.
Given Data / Assumptions:
Concept / Approach:NAND gates are universal and particularly convenient for SOP. A product term can be created with a NAND followed by inversion; the final sum operation (OR) can be formed by a NAND whose inputs are the NANDed product terms with appropriate bubble placement. This yields a standard two-level NAND–NAND structure equivalent to AND–OR, but built entirely from NANDs.
Step-by-Step Solution:
Start with F = P1 + P2 + ... + Pn, where each Pi is an AND of literals.Realize each Pi using a NAND. The output is the complement of Pi (call it Pi’).Feed all Pi’ signals into a final NAND. By De Morgan, NAND of complements = OR of originals.Thus F is implemented with only NAND gates (two-level NAND–NAND network).Verification / Alternative check:Compare to a NOR–NOR approach. NOR is ideal for POS (product-of-sums) implementations. For SOP, NAND–NAND aligns naturally without extra inversion overhead.
Why Other Options Are Wrong:
NOR: Best matches POS, not SOP, for two-level realizations.AND: Not universal alone; needs separate OR and NOT for full functionality.DOOR: Not a logic gate; a distractor.Common Pitfalls:Mistaking NOR as equally suitable for SOP; forgetting bubble pushing when converting AND–OR to NAND–NAND; assuming inverters are always needed rather than using intrinsic NAND inversions.
Final Answer:NAND
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