Skip to main content

Positive XOR equals negative XNOR

Short answer

Positive logic assigns high voltage as 1, low as 0; negative logic reverses this. The same voltage table that gives XOR in positive logic yields XNOR in negative logic because the function becomes complement of XOR with complemented inputs, resulting in A⊙B.

Define positive and negative logic and prove that positive X-OR is equivalent to negative X-NOR.

Positive and negative logic

  • Positive logic: the higher voltage level represents logic 1 and the lower level represents logic 0. Example in TTL: +5 V = 1, 0 V = 0.
  • Negative logic: the higher voltage level represents logic 0 and the lower level represents logic 1. Example: +5 V = 0, 0 V = 1 (or in a –5 V system, –5 V = 1, 0 V = 0).

The same physical circuit gives a different logic function depending on which convention is used. Changing from positive to negative logic means complementing every input and output bit.

Proof: positive XOR = negative XNOR

Take a gate working as XOR in positive logic. Its voltage table (H = high, L = low):

ABY
LLL
LHH
HLH
HHL

Positive logic (H = 1, L = 0):

ABY
000
011
101
110

This is XOR: Y=A⊕BY = A \oplus B.

Negative logic (H = 0, L = 1) for the same circuit:

ABY
111
100
010
001

Output is 1 when both inputs are equal, which is XNOR: Y=A⊕B‾Y = \overline{A \oplus B}.

Algebraic proof: in negative logic the function becomes the complement of positive XOR with complemented inputs:

Yneg=Aˉ⊕Bˉ‾=AˉB+ABˉ‾=A⊕B‾=A⊙B\begin{aligned} Y_{neg} &= \overline{\bar A \oplus \bar B} \\ &= \overline{\bar A B + A \bar B} \\ &= \overline{A \oplus B} = A \odot B \end{aligned}

(since Aˉ⊕Bˉ=A⊕B\bar A \oplus \bar B = A \oplus B). Hence the positive-logic XOR gate is the same circuit as a negative-logic XNOR gate.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Digital Logic chapter 2 questions →