Find an expression for each of these outputs, while minimizing the overall total number of gates. Count the total number of gates. You can assume that all inputs are available in positive and negated forms. You can assume that multi-input gates count as 1 gate each. ab 00 01 11 10 ab 00 cd 01 11 10 ab 00 01 11 10 cd cd 00 1 00 1 00 01 01 1 01 1 1 11 1 1 1 1 11 1 1 11 1 1 1 10 10 1 1 10 F1 F2 F3
Find an expression for each of these outputs, while minimizing the overall total number of gates. Count the total number of gates. You can assume that all inputs are available in positive and negated forms. You can assume that multi-input gates count as 1 gate each. ab 00 01 11 10 ab 00 cd 01 11 10 ab 00 01 11 10 cd cd 00 1 00 1 00 01 01 1 01 1 1 11 1 1 1 1 11 1 1 11 1 1 1 10 10 1 1 10 F1 F2 F3
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Transcribed Image Text:Find an expression for each of these outputs, while minimizing the overall total number of gates. Count
the total number of gates. You can assume that all inputs are available in positive and negated forms.
You can assume that multi-input gates count as 1 gate each.
ab
00
ab
00
ab
00
01
11
10
01
11
10
01
11
10
cd
cd
cd
00
1
00
1
00
01
1
01
1
01
1
11
1
1
1
1
11
1
1
11
1
1
1
10
10
1
1
10
F1
F2
F3
Transcribed Image Text:Using K-Maps, identify hazards in the expression:
f(a, b, c, d) = (a + d')(b' + c + d)(a' + c' + d')(b' + c + d)
Create an equivalent expression that is free of hazards.
Expert Solution
Step 1
For F1 The K-Map can be rounded as follows because we can only round either horizontally rectangles or vertically rectangles of dimensions 1x2, 2x1, 1x4, 4x1, 2x4, 4x2 or squares of dimensions 1x1, 2x2, 4x4 for 4 input K-map.
So the minimized function of this becomes ( (a'+a) (b'+b)cd ) + (ab')(c')( d+ d')
= cd + ab'c'
Number of gates required
cd, ab'c' -> 2 AND gate
cd + ab'c' -> 1 OR gate
Total 3 gates
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