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Positive/negative logic; Ex-NOR using NOR

Short answer

Positive logic uses HIGH=1, LOW=0; negative logic uses HIGH=0, LOW=1. The same physical gate becomes AND in positive logic but OR in negative logic. Ex-NOR (XNOR) can be built with four NOR gates: compute (A+B)′, then (A+(A+B)′)′ and (B+(A+B)′)′, and NOR these to get A⊙B.

Describe positive and negative logic with an example. Construct Ex-NOR gate using only NOR gates.

Positive and negative logic

  • Positive logic: HIGH voltage = 1, LOW voltage = 0. Example: TTL with +5 V = 1, 0 V = 0.
  • Negative logic: HIGH voltage = 0, LOW voltage = 1. Example: +5 V = 0, 0 V = 1.

Example: a gate gives a HIGH output only when both inputs are HIGH (L L → L, L H → L, H L → L, H H → H).

  • In positive logic this is 00→0, 01→0, 10→0, 11→1: an AND gate.
  • In negative logic this is 11→1, 10→1, 01→1, 00→0: an OR gate.

So positive AND = negative OR (and positive NAND = negative NOR, positive XOR = negative XNOR).

Ex-NOR gate using only NOR gates

Y=A⊙B=AB+AˉBˉ=ABˉ+AˉB‾Y = A \odot B = AB + \bar A \bar B = \overline{A\bar B + \bar A B} needs four 2-input NOR gates (the dual of the 4-NAND XOR).

G1=A+B‾G2=A+G1‾=Aˉ(A+B)=AˉBG3=B+G1‾=Bˉ(A+B)=ABˉY=G2+G3‾=AˉB+ABˉ‾=A⊙B\begin{aligned} G_1 &= \overline{A + B} \\ G_2 &= \overline{A + G_1} = \bar A (A + B) = \bar A B \\ G_3 &= \overline{B + G_1} = \bar B (A + B) = A \bar B \\ Y &= \overline{G_2 + G_3} = \overline{\bar A B + A \bar B} = A \odot B \end{aligned} 
 A --+---------------[NOR G2]--+
     |                 |       |
     +--[NOR G1]---G1--+       +--[NOR G4]-- Y
     |                 |       |
 B --+---------------[NOR G3]--+
ABG1G_1G2G_2G3G_3Y
001001
010100
100010
110001

✓ Answer checked by Claude Opus 5.5 (high effort)

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