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Q6E

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Discrete Mathematics and its Applications
Found in: Page 841
Discrete Mathematics and its Applications

Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

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Short Answer

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

a)

b)

c)

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.

See the step by step solution

Step by Step Solution

Step 1: 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

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

Step 2: 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

Step 3: 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:

Step 4: 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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