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Use \(K{\bf{ - }}\)maps to find simpler circuits with the same output as each of the circuits shown.

a)

b)

c)

Short Answer

Expert verified

a) The output of the given K-map is

b) The output of the given K-map is

c) Use the zero-property and the output is0.

Step by step solution

01

Definition

Recall the following laws:

Idempotent law:

\(\begin{array}{c}{\bf{x + x = x}}\\x \cdot x{\bf{ = x}}\end{array}\)

Commutative law:

\(\begin{array}{c}x{\bf{ + y = y + x}}\\{\bf{xy = yx}}\end{array}\)

Distributive law:

\(\begin{array}{c}{\bf{x + yz = }}\left( {{\bf{x + y}}} \right)\left( {{\bf{x + z}}} \right)\\{\bf{x}}\left( {{\bf{y + z}}} \right){\bf{ = xy + xz}}\end{array}\)

Zero property:

\({\bf{x\bar x = 0}}\)

Domination law

\(\begin{array}{c}{\bf{x + 1 = 1}}\\x \cdot 0 = 0\end{array}\)

Identity law

\(\begin{array}{c}{\bf{x + 0 = x}}\\x \cdot 1{\bf{ = x}}\end{array}\)

02

AND Gate

A \(K{\bf{ - }}\)map for a function in three variables is a table with four columns \(yz,y\bar z,\bar y\bar z\) and \(\bar yz\); which contains all possible combinations of \(y\) and \(z\) and two rows \(x\) and \(\bar x\).

Place a \(1\) in the cell corresponding to each term in the given sum \(xyz + \bar xyz\).

\(xyz\): place a \(1\) in the cell corresponding to row \(x\) and column \(yz\)

\(\bar xyz\): place a \(1\) in the cell corresponding to row \(\bar x\) and column\(yz\)

Note that all \(1\)'s occur in the column \(yz\), which implies that the simplest Boolean expression is \(yz\).

Please write all the mathematical equations/expressions/terms of MATHTYPE in BLACK ( Not bold) in the solution part.

Done

The Boolean product is represented by an \(AND\) gate, thus a simpler circuit is then:

Please write all the mathematical equations/expressions/terms of MATHTYPE in BLACK ( Not bold) in the solution part.

Done

03

Inverter

A \(K - \)map for a function in three variables is a table with four columns \(yz,y\bar z,\bar y\bar z\) and \(\bar yz\); which contains all possible combinations of \(y\) and \(z\) and two rows \(x\) and \(\bar x\).

Place a \(1\) in the cell corresponding to each term in the given sum \(xy\bar z + x\bar y\bar z + \bar xy\bar z + \bar x\bar y\bar z\)

\(xy\bar z\): place a \(1\) in the cell corresponding to row \(x\) and column \(y\bar z\)

\(x\bar y\bar z\): place a \(1\) in the cell corresponding to row \(x\) and column \(\bar y\bar z\)

\(\bar xy\bar z\): place a \(1\) in the cell corresponding to row \(\bar x\) and column \(y\bar z\)

\(\bar x\bar y\bar z\): place a \(1\) in the cell corresponding to row \(\bar x\) and column\(\bar y\bar z\)

Note that all \(1\)'s occur in the columns \(y\bar z\) and \(\bar y\bar z\) (which are all columns containing \(\bar z\), which implies that the simplest Boolean expression is \(\bar z\).

Please write all the mathematical equations/expressions/terms of MATHTYPE in BLACK ( Not bold) in the solution part.

CORRECT THROUGHOUT THE FILE

Done

The complement is represented by an inverter; thus, a simpler circuit is then:

04

Inverter and the Boolean product

Simplify the given expression using logical equivalences.Using the commutative, idempotent, distributive, domination law and zero property.

\(\begin{array}{c}\bar xyz\left( {\left( {x + \bar z} \right) + \left( {\bar y + \bar z} \right)} \right) = \bar xyz\left( {x + \bar z + \bar y + \bar z} \right)\\ = \bar xyz\left( {x + \bar y + \bar z + \bar z} \right)\\ = \bar xyz\left( {x + \bar y + \bar z} \right)\\ = x\bar xyz + \bar xy\bar yz + \bar xyz\bar z\end{array}\)

\(\begin{array}{c} = 0 \cdot yz + 0 \cdot \bar xz + 0 \cdot \bar xy\\ = 0 + 0 + 0\\ = 0\end{array}\)

The simplest Boolean expression is \(0\). However, we cannot express this with a logic gate (if we actually want to use a logic gate), but then use the zero property \(x\bar x = 0\).The complement is represented by an inverter and the Boolean product by an \(AND\) gate, thus a simpler circuit is then:

Please write all the mathematical equations/expressions/terms of MATHTYPE in BLACK ( Not bold) in the solution part.

CORRECT THROUGHOUT THE FILE

Done

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Most popular questions from this chapter

Let \({\bf{x}}\) and \({\bf{y}}\) belong to \(\left\{ {{\bf{0,1}}} \right\}\). Does it necessarily follow that \({\bf{x = y}}\) if there exists a value \({\bf{z}}\) in \(\left\{ {{\bf{0,1}}} \right\}\) such that,

\(\begin{array}{l}{\bf{a) xz = yz?}}\\{\bf{b) x + z = y + z?}}\\{\bf{c) x}} \oplus {\bf{z = y}} \oplus {\bf{z?}}\\{\bf{d) x}} \downarrow {\bf{z = y}} \downarrow {\bf{z?}}\\{\bf{e) x}}|{\bf{z = y}}|z{\bf{?}}\end{array}\)

A Boolean function \({\bf{F}}\) is called self-dual if and only if \({\bf{F}}\left( {{{\bf{x}}_{\bf{1}}}{\bf{, \ldots ,}}{{\bf{x}}_{\bf{n}}}} \right){\bf{ = }}\overline {{\bf{F}}\left( {{{{\bf{\bar x}}}_{\bf{1}}}{\bf{, \ldots ,}}{{{\bf{\bar x}}}_{\bf{n}}}} \right)} \).

How many different Boolean functions \({\bf{F(x,y,z)}}\) are there such that \({\bf{F(\bar x,y,z) = F(x,\bar y,z) = F(x,y,\bar z)}}\) for all values of the Boolean variables \({\bf{x,y}}\), and \({\bf{z}}\)\({\bf{?}}\)

Use a \({\bf{K}}\)-map to find a minimal expansion as a Boolean sum of Boolean products of each of these functions in the variables \({\bf{x,y}}\), and \({\bf{z}}\).

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

Simplify these expressions.

\(\begin{array}{l}{\bf{a) x}} \oplus {\bf{0}}\\{\bf{b) x}} \oplus {\bf{1}}\\{\bf{c) x}} \oplus {\bf{x}}\\{\bf{d) x}} \oplus {\bf{\bar x}}\end{array}\)

Construct a half adder using \(OR\) gates, \(AND\) gates, and inverters.

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