Question

Given four inputs: a, b, c & d, where (a, b) represents a 2-bit unsigned binary number X; and (c, d) represents a 2-bit unsigned binary number Y (i.e. both X and Y are in the range #0 to #3). The output is z, which is 1 whenever X > Y, and 0 otherwise (this circuit is part of a "2-bit comparator"). For instance, if a = 1, b = 0 (i.e. X = b10 => #2); c = 0, d = 1 (i.e. Y = b01 => #1); then z = 1, since b10 > b01
Just need truth table, and boolean expression (simplified) for these thumbs up.

Answer 1

Answer:

z = a.c' + a.b.d' + b.c'.d'

Explanation:

The truth table for this question is provided in the attachment to this question.

N.B - a' = not a!

The rows with output of 1 come from the following relations: 01 > 00, 10 > 00, 10 > 01, 11 > 00, 11 > 01, 11 > 10

This means that the Boolean expression is a sum of all the rows with output of 1.

z = a'bc'd' + ab'c'd' + ab'c'd + abc'd' + abc'd + abcd'

On simplification,

z = bc'd' + ab'c' + ac'd' + ac'd + abc' + abd'

z = ac' + abd' + bc'd'

Hope this helps!