Essentials of Computer Organization and Architecture
Essentials of Computer Organization and Architecture
5th Edition
ISBN: 9781284123036
Author: Linda Null
Publisher: Jones & Bartlett Learning
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Chapter 3, Problem 36E
Program Plan Intro

a.

K-Map:

  • K-Map stands for Karnaugh Map which is used to reduce the logic functions more easily and quickly.
  • By using K-Map, the Boolean expressions with two to four variables are easily reduced.

Expert Solution
Check Mark

Explanation of Solution

Simplification of K-Map using Boolean identities:

 Given:

Essentials of Computer Organization and Architecture, Chapter 3, Problem 36E , additional homework tip  1

 Solution:

Essentials of Computer Organization and Architecture, Chapter 3, Problem 36E , additional homework tip  2

 The simplified expression using K-Map is F=y' + z' + x

 The following steps are used to obtain the Boolean expressions using Boolean identities.

 Step1:

 The group the 1’s in the table. Then write the Boolean expression according to the mapping. Write 0 term’s as complemented variable like x’ and 1 term’s as it is like x. Here “X” variable denotes don’t care which means the user will take the “X” value as either 0 or 1.

Essentials of Computer Organization and Architecture, Chapter 3, Problem 36E , additional homework tip  3

 The below expression is obtained from the K-Map

 F = x'y'z' + x'y'z + xy'z' + xy'z= x'y'(z' + z)+ xy'(z' + z)                            (by Distribution law)=x'y'(1)+ xy'(1)                                           (by Inverse law)=x'y'+ xy'                                                        (by Identity law)

    (x' + x)y'                                                        (by Distribution law)(1)y'                                                               (by Inverse law)= y'                                                                      (by Identity law)

 Step2:

 The group the 1’s in the table. Then write the Boolean expression according to the mapping. Write 0 term’s as complemented variable like x’ and 1 term’s as it is like x. Here “X” variable denotes don’t care which means the user will take the “X” value as either 0 or 1.

Essentials of Computer Organization and Architecture, Chapter 3, Problem 36E , additional homework tip  4

 The below expression is obtained from the K-Map

 F = x'y'z' + xy'z' + x'yz' + xyz'= y'z'(x' + x)+ yz'(x' + x)                            (by Distribution law)=y'z'(1)+ yz'(1)                                           (by Inverse law)=y'z'+ yz'                                                        (by Identity law)

    (y' + y)z'                                                        (by Distribution law)(1)z'                                                               (by Inverse law)= z'                                                                      (by Identity law)

 Step3:

 The group the 1’s in the table. Then write the Boolean expression according to the mapping. Write 0 term’s as complemented variable like x’ and 1 term’s as it is like x. Here “X” variable denotes don’t care which means the user will take the “X” value as either 0 or 1.

Essentials of Computer Organization and Architecture, Chapter 3, Problem 36E , additional homework tip  5

 The below expression is obtained from the K-Map

 F = xy'z' + xy'z + xyz + xyz'= xy'(z' + z)+ xy(z' + z)                            (by Distribution law)=xy'(1)+ xy(1)                                           (by Inverse law)=xy'+ xy                                                        (by Identity law)

    (y' + y)x                                                        (by Distribution law)(1)x                                                               (by Inverse law)= x                                                                      (by Identity law)

 Step4:

 Group all the expressions

 F=y' + z' + x

Therefore, the simplified expression using Boolean identities is “F=y' + z' + x”.

Program Plan Intro

b.

K-Map:

  • K-Map stands for Karnaugh Map which is used to reduce the logic functions more easily and quickly.
  • By using K-Map, the Boolean expressions with two to four variables are easily reduced.

Expert Solution
Check Mark

Explanation of Solution

Simplification of K-Map using Boolean identities:

 Given:

Essentials of Computer Organization and Architecture, Chapter 3, Problem 36E , additional homework tip  6

 Solution:

Essentials of Computer Organization and Architecture, Chapter 3, Problem 36E , additional homework tip  7

 The simplified expression using K-Map is F=w'x' + x'z' + w'z + w'y + yz

 The following steps are used to obtain the Boolean expressions using Boolean identities.

 Step1:

 The group the 1’s in the table. Then write the Boolean expression according to the mapping. Write 0 term’s as complemented variable like x’ and 1 term’s as it is like x. Here “X” variable denotes don’t care which means the user will take the “X” value as either 0 or 1.

Essentials of Computer Organization and Architecture, Chapter 3, Problem 36E , additional homework tip  8

 The below expression is obtained from the K-Map

 F = w'x'y'z' + w'x'y'z + w'x'yz + w'x'yz'= w'x'y'(z' + z)+ w'x'y(z' + z)                            (by Distribution law)=w'x'y'(1)+ w'x'y(1)                                           (by Inverse law)=w'x'y'+ w'x'y                                                       (by Identity law)

    = w'x'(y' + y)                                                             (by Distribution law)= w'x'(1)                                                                     (by Inverse law)= w'x'                                                                           (by Identity law)

 Step2:

 The group the 1’s in the table. Then write the Boolean expression according to the mapping. Write 0 term’s as complemented variable like x’ and 1 term’s as it is like x. Here “X” variable denotes don’t care which means the user will take the “X” value as either 0 or 1.

Essentials of Computer Organization and Architecture, Chapter 3, Problem 36E , additional homework tip  9

 The below expression is obtained from the K-Map

 F = w'x'y'z' + w'x'yz'+ wx'y'z' + wx'yz' = w'x'z'(y' + y)+ wx'z'(y' + y)                            (by Distribution law)=w'x'z'(1)+ wx'z'(1)                                           (by Inverse law)=w'x'z'+ wx'z'                                                       (by Identity law)

    = x'z'(w' + w)                                                             (by Distribution law)= x'z'(1)                                                                     (by Inverse law)= x'z'                                                                           (by Identity law)

 Step3:

 The group the 1’s in the table. Then write the Boolean expression according to the mapping. Write 0 term’s as complemented variable like x’ and 1 term’s as it is like x. Here “X” variable denotes don’t care which means the user will take the “X” value as either 0 or 1.

Essentials of Computer Organization and Architecture, Chapter 3, Problem 36E , additional homework tip  10

 The below expression is obtained from the K-Map

 F = w'x'y'z + w'x'yz + w'xy'z + w'xyz= w'x'z(y' + y)+ w'xz(y' + y)                            (by Distribution law)=w'x'z(1)+ w'xz(1)                                           (by Inverse law)=w'x'z+ w'xz                                                       (by Identity law)

    = w'z(x' + x)                                                             (by Distribution law)= w'z(1)                                                                     (by Inverse law)= w'z                                                                           (by Identity law)

 Step4:

 The group the 1’s in the table. Then write the Boolean expression according to the mapping. Write 0 term’s as complemented variable like x’ and 1 term’s as it is like x. Here “X” variable denotes don’t care which means the user will take the “X” value as either 0 or 1.

Essentials of Computer Organization and Architecture, Chapter 3, Problem 36E , additional homework tip  11

 The below expression is obtained from the K-Map

 F = w'xyz + w'xyz' + w'x'yz + w'x'yz'= w'xy(z' + z)+ w'x'y(z' + z)                            (by Distribution law)=w'xy(1)+ w'x'y(1)                                           (by Inverse law)=w'xy+ w'x'y                                                       (by Identity law)

    = w'y(x' + x)                                                             (by Distribution law)= w'y(1)                                                                     (by Inverse law)= w'y                                                                           (by Identity law)

 Step5:

 The group the 1’s in the table. Then write the Boolean expression according to the mapping. Write 0 term’s as complemented variable like x’ and 1 term’s as it is like x. Here “X” variable denotes don’t care which means the user will take the “X” value as either 0 or 1.

Essentials of Computer Organization and Architecture, Chapter 3, Problem 36E , additional homework tip  12

 The below expression is obtained from the K-Map

 F = w'x'yz + w'xyz + wxyz + wx'yz= w'yz(x' + x)+ wyz(x' + x)                            (by Distribution law)=w'yz(1)+ wyz(1)                                           (by Inverse law)=w'yz+ wyz                                                       (by Identity law)

    = yz(w' + w)                                                           (by Distribution law)= yz(1)                                                                    (by Inverse law)= yz                                                                           (by Identity law)

 Step6:

 Group all the expressions

 F=w'x' + x'z' + w'z + w'y + yz

Therefore, the simplified expression using Boolean identities is “F=w'x' + x'z' + w'z + w'y + yz”.

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Chapter 3 Solutions

Essentials of Computer Organization and Architecture

Ch. 3 - Prob. 11RETCCh. 3 - Prob. 12RETCCh. 3 - Prob. 13RETCCh. 3 - Prob. 14RETCCh. 3 - Prob. 15RETCCh. 3 - Prob. 16RETCCh. 3 - Prob. 17RETCCh. 3 - Prob. 18RETCCh. 3 - Prob. 19RETCCh. 3 - Prob. 20RETCCh. 3 - Prob. 21RETCCh. 3 - Prob. 22RETCCh. 3 - Prob. 23RETCCh. 3 - Prob. 24RETCCh. 3 - Prob. 25RETCCh. 3 - Prob. 26RETCCh. 3 - Prob. 1ECh. 3 - Prob. 2ECh. 3 - Prob. 3ECh. 3 - Prob. 4ECh. 3 - Prob. 5ECh. 3 - Prob. 6ECh. 3 - Prob. 7ECh. 3 - Prob. 8ECh. 3 - Prob. 9ECh. 3 - Prob. 10ECh. 3 - Prob. 11ECh. 3 - Prob. 12ECh. 3 - Prob. 13ECh. 3 - Prob. 14ECh. 3 - Prob. 15ECh. 3 - Prob. 16ECh. 3 - Prob. 17ECh. 3 - Prob. 18ECh. 3 - Prob. 19ECh. 3 - Prob. 20ECh. 3 - Prob. 21ECh. 3 - Prob. 22ECh. 3 - Prob. 23ECh. 3 - Prob. 24ECh. 3 - Prob. 25ECh. 3 - Prob. 26ECh. 3 - Prob. 27ECh. 3 - Prob. 28ECh. 3 - Prob. 29ECh. 3 - Prob. 30ECh. 3 - Prob. 31ECh. 3 - Prob. 32ECh. 3 - Prob. 33ECh. 3 - Prob. 34ECh. 3 - Prob. 35ECh. 3 - Prob. 36ECh. 3 - Prob. 37ECh. 3 - Prob. 38ECh. 3 - Prob. 39ECh. 3 - Prob. 40ECh. 3 - Prob. 41ECh. 3 - Prob. 42ECh. 3 - Prob. 43ECh. 3 - Prob. 44ECh. 3 - Prob. 45ECh. 3 - Prob. 46ECh. 3 - Prob. 47ECh. 3 - Prob. 48ECh. 3 - Prob. 49ECh. 3 - Prob. 50ECh. 3 - Prob. 51ECh. 3 - Prob. 52ECh. 3 - Prob. 53ECh. 3 - Prob. 54ECh. 3 - Prob. 55ECh. 3 - Prob. 56ECh. 3 - Prob. 57ECh. 3 - Prob. 58ECh. 3 - Prob. 59ECh. 3 - Prob. 60ECh. 3 - Prob. 61ECh. 3 - Prob. 62ECh. 3 - Prob. 63ECh. 3 - Prob. 64ECh. 3 - Prob. 65ECh. 3 - Prob. 66ECh. 3 - Prob. 67ECh. 3 - Prob. 68ECh. 3 - Prob. 70ECh. 3 - Prob. 71ECh. 3 - Prob. 72ECh. 3 - Prob. 73ECh. 3 - Prob. 74ECh. 3 - Prob. 75ECh. 3 - Prob. 76ECh. 3 - Prob. 77ECh. 3 - Prob. 78ECh. 3 - Prob. 79ECh. 3 - Prob. 80E
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