Question

Find the sum-of-products expansions of these Boolean functions.

a)\({\bf{F}}\left( {{\bf{x}},{\rm{ }}{\bf{y}}} \right) = \overline {\bf{x}} {\bf{ + y}}\)

b)\({\bf{F}}\left( {{\bf{x}},{\rm{ }}{\bf{y}}} \right) = {\bf{x}}\overline {\bf{y}} \)

c)\({\bf{F}}\left( {{\bf{x}},{\rm{ }}{\bf{y}}} \right) = {\bf{1}}\)

d)\({\bf{F}}\left( {{\bf{x}},{\rm{ }}{\bf{y}}} \right) = \overline {\bf{y}} \)

Short answer

The sum of product expansions is

(a) The sum of product is \(xy + \overline x y + \overline x \overline y \).

(b) The sum of product is \(x\overline y \).

(c) The sum of product is \(xy + \overline x y + x\overline y + \overline x \overline y \).

(d) The sum of product is \(x\overline y + \overline x \overline y \).

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.

(a) Finding the solution.

Here \(F\left( {x,{\rm{ }}y} \right) = \overline x + y\).

To solve this use Boolean identities.

\(\begin{aligned}F\left( {x,y} \right) &= \overline x + y\\ &= \overline x .1 + y.1\\ &= \overline x \left( {y + \overline y } \right) + y\left( {x + \overline x } \right)\\ &= \overline x y + \overline x \overline y + yx + y\overline x\}\\\ &= \overline x y + \overline x \overline y + xy + y\overline x \\ &= \overline x y + \overline x \overline y + xy + \overline x y \\&= xy + \overline x y + \overline x \overline y \end{aligned}\)

(b) Discover the result.

Here \(F\left( {x,{\rm{ }}y} \right) = x\overline y \)

It is already in the sum of products form. Thus, the sum of products expansion of \(x\overline y \) is \(x\overline y \) itself.

(c) Evaluate the result.

Here \(F\left( {x,{\rm{ }}y} \right) = 1\)

To solve this use Boolean identities.

\(\begin{aligned}F\left( {x,y} \right) &= 1\\ &= 1.1\\ &= \left( {x + \overline x } \right)\left( {y + \overline y } \right)\\ &= \left( {x + \overline x } \right)y + \left( {x + \overline x } \right)y\\ &= y\left( {x + \overline x } \right) + \overline y \left( {x + \overline x } \right)\\ &= yx + y\overline x + y\overline x + \overline y \overline x \\ &=xy + \overline x y + x\overline y + \overline x \overline y \end{aligned}\)

(d) Solve to get the result.

Here \(F\left( {x,{\rm{ }}y} \right) = \overline y \)

To solve this use Boolean identities.

\(\begin{aligned}F\left( {x,y} \right) &= \overline y \\ &= \overline y .1\\ &= \overline y \left( {x + \overline x } \right)\\ &= \overline y x + \overline y \overline x \\ &= x\overline y + \overline x \overline y \end{aligned}\)

Therefore, by using the Boolean concept get the results.