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DeMorgan’s Theorem-Digital Logic Design-Solution Manual

Exercises, Digital Logic Design

Post: July 20th, 2012
Description
This is solution manual for Digital Logic Design course. It was helpful for assignment Dr. Archan Singh gave us at Punjab Engineering College. It includes: Verification, DeMorgan, Theorem, Second, Distributive, Law, NAND, Gate, Design
This is solution manual for Digital Logic Design course. It was helpful for assignment Dr. Archan Singh gave us at Punjab Engineering College. It includes: Verification, DeMorgan, Theorem, Second, Distributive, Law, NAND, Gate, Design
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XYZ = X + Y + Z Verification of DeMorgan’s Theorem X 0 0 0 0 1 1 1 1 Y 0 0 1 1 0 0 1 1 Z 0 1 0 1 0 1 0 1 XYZ 0 0 0 0 0 0 0 1 XYZ 1 1 1 1 1 1 1 0 X+Y+Z 1 1 1 1 1 1 1 0 X + YZ = ( X + Y ) ⋅ ( X + Z ) The Second Distributive Law X 0 0 0 0 1 1 1 1 Y 0 0 1 1 0 0 1 1 Z 0 1 0 1 0 1 0 1 YZ 0 0 0 1 0 0 0 1 X+YZ 0 0 0 1 1 1 1 1 X+Y 0 0 1 1 1 1 1 1 X+Z 0 1 0 1 1 1 1 1 (X+Y)(X+Z) 0 0 0 1 1 1 1 1 XY + YZ + XZ = XY + YZ + XZ X 0 0 0 0 1 1 1 1 Y 0 0 1 1 0 0 1 1 Z 0 1 0 1 0 1 0 1 XY 0 0 1 1 0 0 0 0 YZ 0 1 0 0 0 1 0 0 XZ XY+YZ+XZ XY 0 0 0 0 0 0 1 0 1 0 1 1 1 1 1 1 0 0 0 0 1 1 0 0 YZ 0 0 1 0 0 0 1 0 XZ XY+YZ+XZ 0 0 1 0 1 0 0 0 0 1 1 1 1 1 1 0 2-2. a) X Y + XY + XY = (XY+ X Y ) + (X Y + XY) = X(Y + Y) + Y(X + X) + =X+Y = X+Y 1 docsity.com Problem Solutions – Chapter 2 b) A B+ B C + AB + B C = 1 = (A B+ AB ) + (B C + B C) = B(A + A) + B(C + C) =B+B =1 c) Y+XZ+XY =Y+XY+XZ = (Y + X)(Y + Y) + X Z =Y+X+XZ = Y + (X + X)(X + Z) =X+Y+Z d) X Y + Y Z + XZ + XY + Y Z = X Y + Y Z(X + X) + XZ + XY + Y Z ..

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