ITPEC FE Morning April 2018 Question 23
ITPEC FE Morning April 2018 — Question 23 of 80
Logic Circuit Simplification — find an equivalent circuit using Boolean algebra.
Original circuit computes: NOT( (A AND B) OR NAND(B, C) )
Step-by-step simplification:
- NAND(B, C) = NOT(B · C)
- Full expression: NOT( (A·B) OR NOT(B·C) )
- Apply De Morgan's: NOT(A·B) AND (B·C)
- Result: NAND(A, B) AND B AND C
This matches (c): NAND(A, B) fed into a 3-input AND gate with B and C.
Why not others:
- (a) — uses OR instead of AND for combining NAND output with B, C
- (b) — applies NOR at the output stage, producing a different function
- (d) — uses NAND at the output with an extra inversion, wrong result
Key rule: Use De Morgan's laws to simplify NOR/NAND chains: NOT(X OR Y) = NOT(X) AND NOT(Y). When stuck, substitute concrete values (e.g. all 1s, all 0s) to quickly eliminate wrong options.
AI-generated — may contain errors
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