1 Draw the logic circuit to realize the Boolean expression as below. Do not simplify. Y = f(A, B,C) = TA+B)(B+C) 2 Simplify the expression above using DeMorgan's theorm 3 Draw the simplified circuit. 4 Simplify the following expression using DeMorgan's theorem (A -B - C)(A + C)(A+T) 3.4 Simplify the following Boolean functions, using Karnaugh maps: (a) F (r, y, z) = E(2, 3, 6, 7) (b) F (A, B, C, D) = E(4, 6, 7, 15) (c) F (A, B, C, D) = X(3,7, 11, 13, 14, 15) (d) F (w, x, y, z) = E(2, 3, 12, 13, 14, 15) %3D
1 Draw the logic circuit to realize the Boolean expression as below. Do not simplify. Y = f(A, B,C) = TA+B)(B+C) 2 Simplify the expression above using DeMorgan's theorm 3 Draw the simplified circuit. 4 Simplify the following expression using DeMorgan's theorem (A -B - C)(A + C)(A+T) 3.4 Simplify the following Boolean functions, using Karnaugh maps: (a) F (r, y, z) = E(2, 3, 6, 7) (b) F (A, B, C, D) = E(4, 6, 7, 15) (c) F (A, B, C, D) = X(3,7, 11, 13, 14, 15) (d) F (w, x, y, z) = E(2, 3, 12, 13, 14, 15) %3D
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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I would like to get help with these exercises to make me understand how K-maps work and can help understanding circuits better.
Transcribed Image Text:1 Draw the logic circuit to realize the Boolean expression as below. Do not simplify.
Y = f(A, B,C) = (A+B)(B+
C)
%3D
2 Simplify the expression above using DeMorgan's theorm
3 Draw the simplified circuit.
4 Simplify the following expression using DeMorgan's theorem
(A ·B · C)(A+C)(A+T)
3.4 Simplify the following Boolean functions, using Karnaugh maps:
(a) F (r, y, z) = E(2, 3, 6, 7)
(c) F (A, B, C, D) = £(3, 7,11, 13, 14, 15) (d)* F(w, x, y, z) = E(2, 3, 12, 13, 14, 15)
(b) F (A, B, C, D) = X(4, 6, 7, 15)
3.7 Simplify the following Boolean expressions, using four-variable maps:
(a) w'z + xz + x'y + wx'z
(b) AD' + B'C'D+ BCD' + BC'D
(c) AB' C+ B'C'D' + BCD + ACD' + A'B'C + A' BC'D
(d) wxy + xz + wx'z + w'x
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