[논리회로] 논리회로 - Chapter 2 PROBLEMS solution

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책은 LOGIC AND COMPUTER DESIGN FUNDAMENTALS 입니다.

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SUBJECT : PROBLEMS solution
COURSE TITLE : LOGIC AND COMPUTER DESIGN FUNDAMENTALS
PROPOSITION Day : Mon Sep 13, 1999
PROFESSOR : JaeYoung Choi
PRESENTER : Computing.
Information Science Univ.
9443036 JaeSoo Jang

CHAPTER 2

본문내용

CHAPTER 2

2-2. Prove the identity of each of the following Boolean equations, using
algebraic manipulation:

2-5. Simplify the following Boolean expressions to a minimum number of
literals:

2-6. Reduce the following Boolean expressions to the indicated number of
literals:

2-8. Find the complement of the following expressions.
－ －
(a) AB + AB
－ －
(b) ( VW + X )Y + Z
－ － －－ － －
(c) WX( YZ + YZ ) + WX( Y + Z )( Y + Z )
－ －－ －－
(d) ( A + B + C )( AB + C )( A + BC )

2-10. For the Boolean functions E and F, as give in the following truth
table:

2-11. Convert the following expressions into sum-of-products and products-
of-sums forms:
－
(a) ( AB + C )( B + CD )
－ －
(c) ( A + BC + CD )( B + EF )

2-12. Draw the logic diagram for the following Boolean expressions. The
diagram should correspond exactly to the equation.
－
(a) BC + AB + ACD
－ － － － － －
(c) WX( Y + Z ) + ( W + YZ )( X + Z )

2-13. Simplify the following Boolean functions by means of a three-variable
map:

(a) F(X, Y, Z) = ∑m(1, 3, 6, 7)
(c) F(A, B, C) = ∑m(0, 1, 2, 4, 6)

2-14. Simplify the following Boolean expressions, using a map:
(a) XZ + YZ + XYZ
(c) AB + AC + BC + ABC

2-16. Simplify the following Boolean functions, using a map:
(a) F(W, X, Y, Z) = ∑m(1, 3, 4, 6, 7, 13, 15)

2-19. Simplify the following Boolean functions by finding all prime implicants
and essential prime implicants and applying the selection rule:
(a) F(A, B, C, D) = BD + ABC + ACD + ABC + ACD
(c) F(W, X, Y, Z) = ∑m(0, 1, 4, 5, 6, 7, 8, 9, 10, 11, 14, 15)

2-20. Simplify the following Boolean fuctions in product-of-sums form:
(a) F(W, X, Y, Z) = ∑m(0, 1, 2, 6, 8, 9, 10, 11, 14, 15)


2-21. Simplify the following expressions in (1) sum-of-products and (2)
product-of-sums forms:
(a) AC + BD + ACD + ABCD
(c) (A + B + D)(A + D)(A + B + D)(A + B + C +D)

2-22. Simplify the following Boolean fuctions F together with the don't-care
conditions d:
(a) F(X, Y, Z) = ∑m(0, 1, 2, 4, 5), d(X, Y, Z) = ∑m(3, 6, 7)
(c) F(A, B, C, D) = ∑m(1, 3, 5, 7, 9, 15), d<font color=aaaaff>..</font>

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