Question

Find the sum-of-products expansions of these Booleanfunctions.

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

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

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

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

Short answer

The sum of products expansions are

(a) The sum of products is\(xyz + x\overline y z + xy\overline z + x\overline y \overline z + \overline x yz + \overline x y\overline z + \overline x \overline y z\).

(b) The sum of product is \(xyz + xy\overline z + \overline x yz\).

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

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

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) Find the result.

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

To solve this use Boolean identities.

\(\begin{aligned}F\left( {x,y} \right) &=x + y + z\\ &= x.1 + y.1 + z.1\\ &= x.1.1 + y.1.1 + z.1.1\\ &= x\left( {y + \overline y } \right)\left( {z + \overline z } \right) + y\left( {x + \overline x } \right)\left( {z + \overline z } \right) + z\left( {x + \overline x } \right)\left( {y + \overline y } \right)\\&= \left( {xy + x\overline y } \right)\left( {z + \overline z } \right) + \left( {yx +y\overline x } \right)\left( {z + \overline z } \right) + \left( {zx + z\overline x } \right)\left( {y + \overline y } \right)\\&= xyz + x\overline y z + xy\overline z + x\overline y \overline z + yxz + y\overline x z + yx\overline z + y\overline x \overline z + zxy + z\overline x y + zx\overline y + z\overline x \overline y \\ &= xyz + x\overline y z + xy\overline z + x\overline y \overline z + xyz + \overline x yz + xy\overline z + \overline x y\overline z + xyz + \overline x yz + x\overline y z + \overline x \overline y z\\ &= xyz + x\overline y z + xy\overline z + x\overline y \overline z + \overline x yz + \overline x y\overline z + \overline x \overline y z\end{aligned}\)

(b) Discover the Solution.

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

To solve this use Boolean identities.

\(\begin{aligned}F\left( {x,y} \right) &= \left( {x + z} \right)y\\ &= xy + zy\\ &= xy.1 + zy.1\\ &= xy\left( {z + \overline z } \right) + zy\left( {x + \overline x } \right)\\ &=xyz + xy\overline z + zyx + zy\overline x \\ &= xyz + xy\overline z + xyz + \overline x yz\\ &= xyz + xy\overline z + \overline x yz\end{aligned}\)

(c) Determine the result.

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

To solve this use Boolean identities.

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

(d) Evaluate the result.

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

To solve this use Boolean identities.

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

Therefore, these are the sum of products of the Boolean functions.