ITPEC FE Morning October 2023 Question 21

NOT on A, then a three-input AND of inverted A, B and C — write the circuit as an expression and simplify it with De Morgan's law.

• •NAND of B and C: B · C
• •NOR of A and that output: NOT (A + B · C)
• •De Morgan turns the NOR into an AND of complements: A AND NOT (B · C)
• •The double negation cancels, leaving A · B · C

So the output is 1 only when A = 0, B = 1 and C = 1. Check: with A = 0, B = C = 1 the NAND gives 0 and the NOR of 0 and 0 gives 1.

Answer (c)

Why not others:

• •(a) — a three-input OR gives A + B + C, which is already 1 for A = B = C = 0, where the original circuit outputs 0
• •(b) — a three-input NAND gives the complement of the correct expression, NOT (A · B · C); it forgets that the double negation cancels
• •(d) — a three-input NOR gives NOT (A + B + C) = A · B · C, true for A = 1, B = C = 0, where the original outputs 0

Key rule: A NOR is an AND of the inverted inputs, and a NAND fed into an inverting input cancels back to a plain AND.

Hardware · AI-written, may contain errors