ITPEC FE Morning October 2020 Question 22

NOT — inverting both inputs of an OR gate gives a NAND gate, by De Morgan's law.

The OR gate outputs Z = X + Y. Put a NOT gate in front of each input, so the OR gate receives X and Y:

• •Output: X + Y
• •De Morgan's law: X + Y = X · Y
• •X · Y is the NAND of X and Y

Check with the truth table: the output is 0 only when X = Y = 1 (both inverted inputs are 0), and 1 otherwise, which is exactly NAND.

Answer (c)

Why not others:

• •(a) AND and (d) OR — fed a single signal, an AND or OR gate passes it through unchanged, so the circuit stays a plain OR gate, X + Y
• •(b) NOR — a NOR gate inverts only when both its inputs are tied to the same signal, when it is simply acting as a NOT gate; used as a two-input gate it combines two signals instead of inverting one, so the gate the conversion needs is NOT

Key rule: De Morgan's laws turn gates into each other by inversion: NOT on every input turns OR into NAND and AND into NOR.

Hardware · AI-written, may contain errors