FUNCIONES BOOLEANAS

Solución:

1. F = A + Ba. F(0,0) = 0 + 0 = 0b. F(0,1) = 0 + 1 = 1c. F(1,0) = 1 + 0 = 1d. F(1,1) = 1 + 1 = 1

2. F = A + Ba. F(0,0) = 0 + 0 = 0 + 1 = 1b. F(0,1) = 0 + 1 = 0 + 0 = 0c. F(1,0) = 1 + 0 = 1 + 1 = 1d. F(1,1) = 1 + 1 = 1 + 0 = 1

3. F = A.B+Ca. F(0,0,0) = 0 . 0 + 0 = 1.0 + 0 = 0b. F(0,0,1) = 0 . 0 + 1 = 1.0 + 1 = 1c. F(0,1,0) = 0 . 1 + 0 = 1.1 + 0 = 1d. F(0,1,1) = 0 . 1 + 1 = 1.1 + 1 = 1e. F(1,0,0) = 1 . 0 + 0 = 0.0 + 0 = 0

f. F(1,0,1) = 1 . 0 + 1 = 0.0 + 1 = 1g. F(1,1,0) = 1 . 1 + 0 = 0.1 + 0 = 0h. F(1,1,1) = 1 . 1 + 1 = 0.1 + 1 = 1

1. F = A + B

A B F0 00 11 01 1

1011

2. G = A.B + A.B

A B F0 00 11 01 1

0101

TABLA 1

PRIMERA FORMA CANONICA

F = AB+AB

SEGUNDA FORMA CANONICA

F = (A+B).(A+B)

TABLA 2

PRIMERA FORMA CANONICA

F = A BC+ A BC+ ABC+A BC

SEGUNDA FORMA CANONICA

F = (A+B+ C)( A+ B + C)( A+B + C)(A+B+C)