ITPEC FE Morning October 2019 Question 24
ITPEC FE Morning October 2019 — Question 24 of 80
NAND-only circuit conversion — replace AND, OR, NOT with NAND gates.
Original circuit: F = NOT(A·B + C)
Step-by-step using De Morgan's law:
- •
NOT(A·B + C) = NOT(A·B) · NOT(C)
NAND equivalents:
- NOT(X) = NAND(X, X)
- AND(X, Y) = NAND(NAND(X,Y), NAND(X,Y))
Exam strategy: When visual parsing of gate diagrams is uncertain, build a truth table for the original function and test each option with sample inputs (e.g., A=1, B=1, C=0 → F=0).
Answer: (b)
Key rule: Any logic circuit can be built using only NAND gates. Use De Morgan's law to decompose, then map each NOT/AND/OR to its NAND equivalent.
AI-generated — may contain errors
The original exam layout is preserved in the image so diagrams, formulas, tables, and code remain accurate.
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