Chapter 3, Problem 36E

a.

Explanation of Solution

Truth table for 1-bit comparator:

Given:

$\begin{array}{l}\text{G=1 if A>B, "0" otherwise}\\ \text{E=1 if A==B, "0" otheriwse}\\ \text{L=1 if A<B, "0" otherwise}\end{array}$

The truth table for 1-bit comparator is as follows,

A	B	G	E	L
0	0	0	1	0
0	1	0	0	1
1	0	1	0	0
1	1	0	1	0

b.

Explanation of Solution

Implementation of “G”, “E”, and “L”:

Implementation of “G”:

• “G” can be implemented by taking the Boolean expression from the truth table where “G” has the output “1”.
• The Boolean expression for “G” is “$\text{G=A}\text{.}\overline{\text{B}}$”.
• The implementation of “G” is as follows,

Implementation of “E”:

• “E” can be implemented by taking the Boolean expression from the truth table where “E” has the output “1”.
• The Boolean expression for “E” is “$\text{E=}\overline{\text{A}}$

c.

Explanation of Solution

Four-bit equality checker:

• To build four-bit equality checker, four 1-bit comparators are used.
• The four-bit equality comparator is constructed as follows,

• The output line “E” of every 1-bit comparator is connected with the input line of AND gate, because it produces the output EQUAL “1”, when “4” inputs of “A” and “B” are equal.
• The output EQUAL produces “0” for the rebuking inputs of “A” and “B”.
• Example:
  • Consider the four-bit numbers “1011” and “1011” and are given to the above circuit.
  • 1-bit from both “A” and “B” are taken by each comparator and the output is generated as follows,
    • “${\text{A}}_0$” is the first bit from “A” and its value is “1”. “${\text{B}}_0$” is the first bit from “B” and its value is “1”. Both bits are equal and the output of “E” for the first comparator is “1”.
    • “${\text{A}}_1$” is the second bit from “A” and its value is “1”