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Chapter 12: Q1E (page 822)

Find a Boolean product of the Boolean variables x, y,and z, or their complements, that has the value 1 if and only if

a)x=y=0, z=1

b)x=0, y=1, z=0

c)x=0, y=z=1

d)x=y=z=0

Short Answer

Expert verified

The Boolean products are

  1. The product is\(\overline x \cdot \overline y \cdot z\).
  2. The product is\(\overline x \cdot y \cdot \overline z \).
  3. The product is\(\overline x yz\).
  4. The product is \(\overline x \overline y \overline z \).

Step by step solution

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01

Definition

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

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

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

02

(a) Find the result.

Here \(x = 0,{\rm{ }}y = 0,{\rm{ }}z = 1\).

If a Boolean variable is 0, the complement of the Boolean variable is 1.

\(\begin{array}{l}\overline x = 1\\\overline y = 0\\z = 1\end{array}\)

The Boolean product of Boolean variable is 1 if all Boolean variable s 1.

\(\overline x \cdot \overline y \cdot z = 1\).

Thus, the Boolean product is \(\overline x \cdot \overline y \cdot z\).

03

(b) Determine the result. 

Here \(x = 0,{\rm{ }}y = 1,{\rm{ }}z = 0\).

If a Boolean variable is 0, the complement of the Boolean variable is 1.

\(\begin{array}{l}\overline x = 1\\y = 1\\\overline z = 1\end{array}\)

The Boolean product of Boolean variable is 1 if all Boolean variable s 1.

\(\overline x \cdot y \cdot \overline z = 1\).

Thus, the Boolean product is \(\overline x y\overline z \).

04

(c) Evaluate the result.

Here \(x = 0,{\rm{ }}y = 1,{\rm{ }}z = 1\).

If a Boolean variable is 0,the complement of the Boolean variable is 1.

\(\begin{array}{c}\overline x = 1\\y = 1\\z = 1\end{array}\)

The Boolean product of Boolean variable is 1 if all Boolean variable s 1.

\(\overline x \cdot y \cdot z = 1\).

Thus, the Boolean product is \(\overline x yz\).

05

(d) Find the result. 

Here \(x = 0,{\rm{ }}y = 0,{\rm{ }}z = 0\).

If a Boolean variable is 0,the complement of the Boolean variable is 1.

\(\begin{array}{c}\overline x = 1\\\overline y = 1\\\overline z = 1\end{array}\)

The Boolean product of Boolean variable is 1 if all Boolean variable s 1.

\(\overline x \cdot \overline y \cdot \overline z = 1\).

Therefore, the Boolean product is \(\overline x \overline y \overline z \).

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Most popular questions from this chapter

Construct a multiplexer using AND gates, OR gates, andinverters that has as input the four bits\({{\bf{x}}_{\bf{o}}}{\bf{,}}{{\bf{x}}_{\bf{1}}}{\bf{,}}{{\bf{x}}_{\bf{2}}}{\bf{,}}{{\bf{x}}_{\bf{3}}}\)and the two control bits\({{\bf{c}}_{\bf{o}}}\)and\({{\bf{c}}_{\bf{1}}}\). Set up the circuit so that\({{\bf{x}}_{\bf{i}}}\)is the output, where iis the value of the two-bit integer\({{\bf{(}}{{\bf{c}}_{\bf{1}}}{{\bf{c}}_{\bf{o}}}{\bf{)}}_{\bf{2}}}\).The depthof a combinatorial circuit can be defined by specifyingthat the depth of the initial input is 0 and if a gate has ndifferent inputs at depths\({{\bf{d}}_{\bf{1}}}{\bf{,}}{{\bf{d}}_{\bf{2}}}{\bf{,}}.....{\bf{,}}{{\bf{d}}_{\bf{n}}}\),respectively, then its outputs have depth equal to max\({\bf{(}}{{\bf{d}}_{\bf{1}}}{\bf{,}}{{\bf{d}}_{\bf{2}}}{\bf{,}}.....{\bf{,}}{{\bf{d}}_{\bf{n}}}{\bf{) + 1}}\); this value is also defined to be the depth of the gate. The depth of a combinatorial circuit is the maximum depth of the gates in the circuit.

Draw the \({\bf{3}}\)-cube \({{\bf{Q}}_{\bf{3}}}\) and label each vertex with the minterm in the Boolean variables \({\bf{x, y}}\), and \({\bf{z}}\) associated with the bit string represented by this vertex. For each literal in these variables indicate the \({\bf{2}}\)-cube \({{\bf{Q}}_{\bf{2}}}\) that is a subgraph of \({{\bf{Q}}_{\bf{3}}}\) and represents this literal.

In Exercises 1–5 find the output of the given circuit.

Use a \(3\)- cube \({{\bf{Q}}_{\bf{3}}}\) to represent each of the Boolean functions in Exercise \(6\) by displaying a black circle at each vertex that corresponds to a \(3\)-tuple where this function has the value \(1\) .

Construct a \({\bf{K}}\)-map for \({\bf{F(x,y,z) = xz + yz + xy\bar z}}{\bf{.}}\) Use this \({\bf{K - }}\)map to find the implicants, prime implicants, and essential prime implicants of \({\bf{F(x,y,z)}}\).
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