Fundamentals of Logic Design 2 Edition Answers to...

2
Fundamentals of Logic Design 2 nd Edition Answers to Chapter 8 Problems 1. 4-bit BCD inputs: A, B, C, D 4-bit excess-3 outputs: W, X, Y, Z W = Σ m (5,6,7,8,9) + d(10-15) X = Σ m (1,2,3,4,9) + d(10-15) Y = Σ m (0,3,4,7,8) + d(10-15) Z = Σ m (0,2,4,6,8) + d(10-15) 0 X X 1 X 2 X X 3 X X 4 X X X A 3 5 X 6 X X B 2 7 X X 8 X X X C 1 9 X X 10 D 0 11 12 13 14 15 W X Y Z 2. W = Σ m (0,4,6,8,11,12,14,15) X = Σ m (0,1,4,5,6,7,8,9,11,12,14,15) Y = Σ m (0,1,4,5,11,15) Z = Σ m (0,1,3,4,5,6,7,8,9,11,12,14,15) 0 X X X X 1 X X X 2 3 X 4 X X X X A 3 5 X X X 6 X X X B 2 7 X X 8 X X X C 1 9 X X 10 D 0 11 X X X X 12 X X X 13 14 X X X 15 X X X X W X Y Z

Transcript of Fundamentals of Logic Design 2 Edition Answers to...

Page 1: Fundamentals of Logic Design 2 Edition Answers to …faculty.uml.edu/dbowden/ClassPages/17-342/Homework/Fundamentals of...Fundamentals of Logic Design 2nd Edition Answers to Chapter

Fundamentals of Logic Design 2nd Edition Answers to Chapter 8 Problems 1. 4-bit BCD inputs: A, B, C, D 4-bit excess-3 outputs: W, X, Y, Z W = Σ m (5,6,7,8,9) + d(10-15) X = Σ m (1,2,3,4,9) + d(10-15) Y = Σ m (0,3,4,7,8) + d(10-15)

Z = Σ m (0,2,4,6,8) + d(10-15) 0 X X 1 X 2 X X 3 X X 4 X X X A 3 5 X 6 X X B 2 7 X X 8 X X X C 1 9 X X 10 D 0 11 12 13 14 15

W X Y Z 2. W = Σ m (0,4,6,8,11,12,14,15) X = Σ m (0,1,4,5,6,7,8,9,11,12,14,15) Y = Σ m (0,1,4,5,11,15)

Z = Σ m (0,1,3,4,5,6,7,8,9,11,12,14,15) 0 X X X X 1 X X X 2 3 X 4 X X X X A 3 5 X X X 6 X X X B 2 7 X X 8 X X X C 1 9 X X 10 D 0 11 X X X X 12 X X X 13 14 X X X 15 X X X X

W X Y Z

Page 2: Fundamentals of Logic Design 2 Edition Answers to …faculty.uml.edu/dbowden/ClassPages/17-342/Homework/Fundamentals of...Fundamentals of Logic Design 2nd Edition Answers to Chapter

3. A B C D

f1 = A’CD + BC’D + ABD’

f2 = A’B’ + A’C’D + ABD’ X X X X X X X X X X X X X X

f3 = B’D’ + A’CD X X X X X X X X X X X

f4 = A’B’ + BD X X X f1 f2 f3 f4

4. A B C D

X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X W X Y Z 5. f1 = A’CD + BC’D + ABD’ f2 = A’B’ + A’C’D + ABD’ f3 = B’D’ + B’C f4 = A’B’ + BD 6. V = A’ + B’C’ + BC (Use two 2-input OR) W = C’D’ + BD’ + ACD (Use one 4-input OR) X = C’D’ + A’C’ + BC + AB’D (Use one 4-input OR) Y = A’C’ + ACD (Use one 2-input OR) Z = X + A’D (Use one 2-input OR)