This document contains 8 algebra of Boolean expressions problems with their step-by-step solutions. The problems involve simplifying Boolean expressions using Boolean algebra rules such as absorption, commutativity, associativity, distributivity, complement, and identity elements.
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Adbequipo4
1. ELECTRONICA DIGITAL - EJERCICIOS ALGEBRA DE BOOLE:
- Gilber Brice単o
- Freddy Chourio
1) F=XYZ+XYZ+ XYZ+ XYZ+ XYZ
F=XY(Z+Z)+ XYZ+ XYZ+ XYZ
F=XY+ XYZ+ XYZ+ XYZ
F=XY+ XYZ+XY(Z+Z)
F=XY+ XYZ+XY
F= XYZ+1
F=1
2) F= ABC + ABC + ABC + ABC + ABC + ABC
F= AC (B+B) + AB (C+C) + ABC + ABC
F= AC + AB + ABC + ABC
F= AC + A (B+ BC) + ABC
F= AC + A (B + C) + ABC
F= C (A + A) + AB + ABC
F= C + AB + ABC
F= C + B (A + AC)
F= C+ B (A+ C)
F= C + BA + BC
3) F = (A+B)(A+C)C
F = (A.A+AC+AB+BC)C
F = A.C.C+ABC+B.C.C
F = AC+ABC+BC
F = AC+BC(A+1)
F = AC+BC
4) F = Y(X+Y).(X+Z)
F = Y.(X.X+XZ+YX+YZ)
F = Y.(XZ+YX+YZ)
F = XYZ+Y.Y.X+Y.Y.Z
F = XYZ
5) F=X(X+Y)+Z+ZY
F=X.X+XY+Z+ZY
F=XY+Z+ZY
F=XY+Z+Y
F=Y(X+1)+Z
F=Y+Z
2. 6) F = AB+AB+AB+AB
F = A(B+B)+A(B+B)
F = A+A
F = 1
7) F= (W+XW+YZ)
F=W.XW.YZ
F=W.(X+W).(Y+Z)
F=W.(XY+XZ+WY+WZ)
F=XYW+XWZ+W.WY+W.WZ
F=XYW+XWZ+WY+WZ
F=XYW+WZ(X+1)+WY
F=XYW+WZ+WY
F=WY(X+1)+WZ
F=WY+WZ
8) F = W[(X+Y)(Z+W)]
F = W + (X+Y) + (Z+W)
F = W + XY + WZ