Chapter 3, Problem 42E

Program Plan Intro

a.

Sum-of-product:

• Sum-of-product is an equation written in the form of AND term for each input combination that produces output as logical 1.
• If the value is 1 write the input variables; otherwise write its complement.
• The OR and AND terms are the final combination of the expression.

Explanation of Solution

Truth table:

The given truth table is as follows:

x	y	z	F1	F2
0	0	0	1	0
0	0	1	1	0
0	1	0	1	1
0	1	1	0	1
1	0	0	0	0
1	0	1	0	0
1	1	0	0	1
1	1	1	0	1

Sum-of-product form of F1:

The sum-of-product form of F1 for the above table is as follows:

$\text{F1}\left(\text{x, y, z}\right)\text{ = x'}\cdot \text{y'}\cdot \text{z' + x'}\cdot \text{y'}\cdot \text{z + x'}\cdot \text{y}\cdot \text{z'}$

Sum-of-product form of F2:

The sum-of-product form of F2 for the above table is as follows:

$\text{F1}\left(\text{x, y, z}\right)\text{ = x'}\cdot \text{y}\cdot \text{z' + x'}\cdot \text{y}\cdot \text{z + x}\cdot \text{y}\cdot \text{z' + x}\cdot \text{y}\cdot \text{z}$

Therefore, the sum-of-product form of F1 and F2 are expressed.

Explanation of Solution

b.

Simplification of F1:

The simplifying the sum-of-product form of F1 is as follows:

$\begin{array}{c}\text{F1}\left(\text{x, y, z}\right)\text{ = x'}\cdot \text{y'}\cdot \text{z' + x'}\cdot \text{y'}\cdot \text{z + x'}\cdot \text{y}\cdot \text{z'}\\ \text{= x'}\cdot \text{y'}\cdot \left(\text{z' + z}\right)\text{ + x'}\cdot \text{y}\cdot \text{z' }\left(\text{by Distributive law}\right)\\ =\text{ x'}\cdot \text{y'}\cdot \left(\text{1}\right)\text{ + x'}\cdot \text{y}\cdot \text{z' }\left(\text{by Inverse law}\right)\\ =\text{ x'}\cdot \text{y' + x'}\cdot \text{y}\cdot \text{z' }\left(\text{by Identity law}\right)\end{array}$

                   $\begin{array}{c}=\text{ x'}\cdot \left(\text{y' + y}\cdot \text{z'}\right)\left(\text{by Distributive law}\right)\\ =\text{ x'}\cdot \left(\text{y' + y' + z}\right)\left(\text{by DeMorgan's law}\right)\\ =\text{ x'}\cdot \left(\text{y' + z}\right)\left(\text{by Idempotent law}\right)\\ =\text{ x'}\cdot \text{y' + x'}\cdot \text{z }\left(\text{by Distributive law}\right)\end{array}$

 Therefore, the simplified Boolean expression F1 is $x\text{'}\cdot y\text{'}+x\text{'}\cdot z$

Sum-of-product form of F2:

The sum-of-product form of F2 for the above table is as follows:

$\begin{array}{c}\text{F1}\left(\text{x, y, z}\right)\text{ = x'}\cdot \text{y}\cdot \text{z' + x'}\cdot \text{y}\cdot \text{z + x}\cdot \text{y}\cdot \text{z' + x}\cdot \text{y}\cdot \text{z}\\ =\text{x'}\cdot \text{y}\cdot \left(\text{z' + z}\right)\text{ + x}\cdot \text{y}\cdot \left(\text{z' + z}\right)\left(\text{by Distributive law}\right)\\ =\text{x'}\cdot \text{y}\cdot \left(\text{1}\right)\text{ + x}\cdot \text{y}\cdot \left(\text{1}\right)\left(\text{by Inverse law}\right)\\ =\text{x'}\cdot \text{y + x}\cdot \text{y }\left(\text{by Identity law}\right)\end{array}$

                   $\begin{array}{c}=\left(\text{x' + x}\right)\cdot \text{y }\left(\text{by Distributive law}\right)\\ =\left(\text{1}\right)\cdot \text{y }\left(\text{by Inverse law}\right)\\ =\text{y }\left(\text{by Identity law}\right)\end{array}$

 Therefore, the simplified Boolean expression F2 is $y$

Explanation of Solution

c.

Circuit diagram:

 The circuit diagram for the F1 and F2 is as follows: