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A threshold gate represents a Boolean function. Find a Boolean expression for the Boolean function represented by this threshold gate.

Short Answer

Expert verified

The Boolean function for the given gate is \({{\bf{x}}_{\bf{3}}}{\bf{ + }}\overline {{{\bf{x}}_{\bf{1}}}} {{\bf{x}}_{\bf{2}}}\).

Step by step solution

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01

Definition

Let \({\bf{B = }}\left\{ {{\bf{0,1}}} \right\}\). Then \({{\bf{B}}^{\bf{n}}}{\bf{ = }}\left\{ {\left( {{{\bf{x}}_{\bf{1}}}{\bf{,}}{{\bf{x}}_{\bf{2}}}{\bf{, \ldots ,}}{{\bf{x}}_{\bf{n}}}} \right)\mid {{\bf{x}}_{\bf{i}}} \in {\bf{B}}} \right.\)for \(\left. {1 \le {\bf{i}} \le {\bf{n}}} \right\}\) is the set of all possible \({\bf{n}}\)-tuples of \({\bf{0's}}\) and \({\bf{1's}}\). The variable \({\bf{x}}\) is called a Boolean variable if it assumes values only from \({\bf{B}}\), that is, if its only possible values are \(0\) and \(1\). A function from \({{\bf{B}}^{\bf{n}}}\) to \({\bf{B}}\) is Boolean function and it has degree \({\bf{n}}\).

02

Using the figure to frame inequality

It needs to figure out which combinations of values for \({{\bf{x}}_{\bf{1}}}{\bf{,}}{{\bf{x}}_{\bf{2}}}\), and \({{\bf{x}}_{\bf{3}}}\) cause the inequality \({\bf{ - }}{{\bf{x}}_{\bf{1}}}{\bf{ + }}{{\bf{x}}_{\bf{2}}}{\bf{ + 2}}{{\bf{x}}_{\bf{3}}} \ge \frac{1}{2}\) to be satisfied. Clearly this will be true if \({{\bf{x}}_{\bf{3}}}{\bf{ = L}}\). If \({{\bf{x}}_{\bf{3}}}{\bf{ = 0}}\), then it will be true if and only if \({{\bf{x}}_{\bf{2}}}{\bf{ = 1}}\) and \({{\bf{x}}_{\bf{1}}}{\bf{ = 0}}\). Therefore, a Boolean expression for this function is \({{\bf{x}}_{\bf{3}}}{\bf{ + }}\overline {{{\bf{x}}_{\bf{1}}}} {{\bf{x}}_{\bf{2}}}\).

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Most popular questions from this chapter

Use a table to express the values of each of these Boolean functions.

\(\begin{array}{l}{\bf{a) F(x,y,z) = \bar z}}\\{\bf{b) F(x,y,z) = \bar xy + \bar yz}}\\{\bf{c) F(x,y,z) = x\bar yz + }}\overline {{\bf{(xyz)}}} \\{\bf{d) F(x,y,z) = \bar y(xz + \bar x\bar z)}}\end{array}\)

Show that you obtain the absorption laws for propositions (in Table \({\bf{6}}\) in Section \({\bf{1}}{\bf{.3}}\)) when you transform the absorption laws for Boolean algebra in Table 6 into logical equivalences.

use the laws in Definition \(1\) to show that the stated properties hold in every Boolean algebra.

Show that in a Boolean algebra, the modular properties hold. That is, show that \({\bf{x}} \wedge {\bf{(y}} \vee {\bf{(x}} \wedge {\bf{z)) = (x}} \wedge {\bf{y)}} \vee {\bf{(x}} \wedge {\bf{z)}}\) and \({\bf{x}} \vee {\bf{(y}} \wedge {\bf{(x}} \vee {\bf{z)) = (x}} \vee {\bf{y)}} \wedge {\bf{(x}} \vee {\bf{z)}}\).

Show that if \({\bf{F}}\) and \({\bf{G}}\) are Boolean functions represented by Boolean expressions in \({\bf{n}}\) variables and \({\bf{F = G}}\), then \({{\bf{F}}^{\bf{d}}}{\bf{ = }}{{\bf{G}}^{\bf{d}}}\), where \({{\bf{F}}^{\bf{d}}}\) and \({{\bf{G}}^{\bf{d}}}\) are the Boolean functions represented by the duals of the Boolean expressions representing \({\bf{F}}\) and \({\bf{G}}\), respectively. (Hint: Use the result of Exercise \(29\).)

Show that these identities hold.

\(\begin{array}{c}a)\;x \oplus y{\bf{ = (x + }}y)\overline {(xy)} \\b)\;x \oplus y{\bf{ = (x\bar y) + }}(\bar xy)\end{array}\)

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