Question

Find the product-of-sums expansion of each of the Boolean functions in Exercise 3.

Short answer

The products of sums expansion of Boolean functions are

(a) The product of sums expansion is \({\bf{F(x,y,z) = x + y + z}}\).

(b) The product of sums expansion is \({\bf{F(x,y,z) = }}\left( {{\bf{x + y + z}}} \right)\left( {{\bf{x + y + }}\overline {\bf{z}} } \right)\left( {{\bf{x + }}\overline {\bf{y}} {\bf{ + z}}} \right)\left( {\overline {\bf{x}} {\bf{ + y + z}}} \right)\left( {\overline {\bf{x}} {\bf{ + y + }}\overline {\bf{z}} } \right)\).

(c) The product of sums expansion is \({\bf{F(x,y,z) = }}\left( {{\bf{x + y + z}}} \right)\left( {{\bf{x + y + }}\overline {\bf{z}} } \right)\left( {{\bf{x + }}\overline {\bf{y}} {\bf{ + z}}} \right)\left( {{\bf{x + }}\overline {\bf{y}} {\bf{ + }}\overline {\bf{z}} } \right)\).

(d) The product of sums expansion is \({\bf{F(x,y,z) = }}\left( {{\bf{x + y + z}}} \right)\left( {{\bf{x + y + }}\overline {\bf{z}} } \right)\left( {{\bf{x + }}\overline {\bf{y}} {\bf{ + z}}} \right)\left( {{\bf{x + }}\overline {\bf{y}} {\bf{ + }}\overline {\bf{z}} } \right)\left( {\overline {\bf{x}} {\bf{ + }}\overline {\bf{y}} {\bf{ + z}}} \right)\left( {\overline {\bf{x}} {\bf{ + }}\overline {\bf{y}} {\bf{ + }}\overline {\bf{z}} } \right)\).

Step-by-step solution

Definition

The complements of an elements\(\overline {\bf{0}} {\bf{ = 1}}\)and\(\overline {\bf{1}} {\bf{ = 0}}\).

The Boolean sum + or OR is 1 if either term is 1.

The Boolean product (.) or AND is 1 if both terms are 1.

Find the product of sums expansion of part (a).(a)

Here, \({\bf{F(x,y,z) = x + y + z}}\)

This is already in the product of some form. Thus, the product of sums expansion of \({\bf{x + y + z}}\) is \({\bf{x + }}\,\,{\bf{y}}\,\,{\bf{ + z}}\).

Determine the product of sums expansion of part (b). (b)

Here, \({\bf{F(x,y,z) = }}\left( {{\bf{x + z}}} \right){\bf{y}}\)

First find the value of F.

X	Y	Z	\(\left( {{\bf{x + z}}} \right)\)	\(\left( {{\bf{x + z}}} \right){\bf{y}}\)	Boolean sum
0	0	0	0	0	\({\bf{x + y + z}}\)
0	0	1	1	0	\({\bf{x + y + }}\overline {\bf{z}} \)
0	1	0	0	0	\({\bf{x + }}\overline {\bf{y}} {\bf{ + z}}\)
0	1	1	1	1	\({\bf{x + }}\overline {\bf{y}} {\bf{ + }}\overline {\bf{z}} \)
1	0	0	1	0	\(\overline {\bf{x}} {\bf{ + y + z}}\)
1	0	1	1	0	\(\overline {\bf{x}} {\bf{ + y + }}\overline {\bf{z}} \)
1	1	0	1	1	\(\overline {\bf{x}} {\bf{ + }}\overline {\bf{y}} {\bf{ + z}}\)
1	1	1	1	1	\(\overline {\bf{x}} {\bf{ + }}\overline {\bf{y}} {\bf{ + }}\overline {\bf{z}} \)

The product of sums expansion F then contains a factor for every row that has a 0 in the column. A factor in the sum of the three variables.

\({\bf{F(x,y,z) = }}\left( {{\bf{x + y + z}}} \right)\left( {{\bf{x + y + }}\overline {\bf{z}} } \right)\left( {{\bf{x + }}\overline {\bf{y}} {\bf{ + z}}} \right)\left( {\overline {\bf{x}} {\bf{ + y + z}}} \right)\left( {\overline {\bf{x}} {\bf{ + y + }}\overline {\bf{z}} } \right)\).

Evaluate the product of sums expansion of part (c). (c)

Here, \({\bf{F(x,y,z) = x}}\).

First find the value of F.

X	Y	Z	\({\bf{F(x,y,z) = x}}\)	Boolean sum
0	0	0	0	\({\bf{x + y + z}}\)
0	0	1	0	\({\bf{x + y + }}\overline {\bf{z}} \)
0	1	0	0	\({\bf{x + }}\overline {\bf{y}} {\bf{ + z}}\)
0	1	1	0	\({\bf{x + }}\overline {\bf{y}} {\bf{ + }}\overline {\bf{z}} \)
1	0	0	1	\(\overline {\bf{x}} {\bf{ + y + z}}\)
1	0	1	1	\(\overline {\bf{x}} {\bf{ + y + }}\overline {\bf{z}} \)
1	1	0	1	\(\overline {\bf{x}} {\bf{ + }}\overline {\bf{y}} {\bf{ + z}}\)
1	1	1	1	\(\overline {\bf{x}} {\bf{ + }}\overline {\bf{y}} {\bf{ + }}\overline {\bf{z}} \)

The product of sums expansion F then contains a factor for every row that has a 0 in the column. A factor in the sum of the three variables.

\({\bf{F(x,y,z) = }}\left( {{\bf{x + y + z}}} \right)\left( {{\bf{x + y + }}\overline {\bf{z}} } \right)\left( {{\bf{x + }}\overline {\bf{y}} {\bf{ + z}}} \right)\left( {{\bf{x + }}\overline {\bf{y}} {\bf{ + }}\overline {\bf{z}} } \right)\).

Evaluate the product of sums expansion for part (d).(d)

Here, \({\bf{F(x,y,z) = x}}\overline {\bf{y}} \)

First find the value of F.

X	Y	Z	\(\overline {\bf{y}} \)	\({\bf{F(x,y,z) = x}}\)	Boolean sum
0	0	0	1	0	\({\bf{x + y + z}}\)
0	0	1	1	0	\({\bf{x + y + }}\overline {\bf{z}} \)
0	1	0	0	0	\({\bf{x + }}\overline {\bf{y}} {\bf{ + z}}\)
0	1	1	0	0	\({\bf{x + }}\overline {\bf{y}} {\bf{ + }}\overline {\bf{z}} \)
1	0	0	1	1	\(\overline {\bf{x}} {\bf{ + y + z}}\)
1	0	1	1	1	\(\overline {\bf{x}} {\bf{ + y + }}\overline {\bf{z}} \)
1	1	0	0	0	\(\overline {\bf{x}} {\bf{ + }}\overline {\bf{y}} {\bf{ + z}}\)
1	1	1	0	0	\(\overline {\bf{x}} {\bf{ + }}\overline {\bf{y}} {\bf{ + }}\overline {\bf{z}} \)

The product of sums expansion F then contains a factor for every row that has a 0 in the column. A factor in the sum of the three variables.

Therefore, the sum of products expansions is \({\bf{F(x,y,z) = }}\left( {{\bf{x + y + z}}} \right)\left( {{\bf{x + y + }}\overline {\bf{z}} } \right)\left( {{\bf{x + }}\overline {\bf{y}} {\bf{ + z}}} \right)\left( {{\bf{x + }}\overline {\bf{y}} {\bf{ + }}\overline {\bf{z}} } \right)\left( {\overline {\bf{x}} {\bf{ + }}\overline {\bf{y}} {\bf{ + z}}} \right)\left( {\overline {\bf{x}} {\bf{ + }}\overline {\bf{y}} {\bf{ + }}\overline {\bf{z}} } \right)\).