ITPEC FE Morning April 2020 Question 22
ITPEC FE Morning April 2020 — Question 22 of 80
NAND Gate Equivalence — simplify a multi-gate circuit using De Morgan's law.
Original circuit: NAND(x, y) → OR ← NOT(z)
Step-by-step:
- •
NAND(x, y)=¬(x·y) - •
NOT(z)=¬z - •
OR(¬(x·y), ¬z)=¬(x·y) + ¬z
Apply De Morgan's law: ¬A + ¬B = ¬(A·B)
- •
¬(x·y) + ¬z=¬((x·y)·z)=¬(x·y·z)
This equals NAND(x, y, z) — a single 3-input AND gate with inverted output.
Why not others:
- (a) AND(x,y,z) — no inversion, gives x·y·z
- (b) AND(¬y,¬z) → OR(x,...) — gives x + ¬y·¬z, different function
- (c) OR(y,z) → NAND(x,...) — gives ¬(x·(y+z)), different function
Key rule: De Morgan's law lets you convert between OR-of-NOTs and NAND: ¬A + ¬B = ¬(A·B).
AI-generated — may contain errors
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