Ejercicios de Funciones Lógicas
Ejercicios de Funciones Lógicas
Ejercicios de Funciones Lógicas
Create successful ePaper yourself
Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.
Por tanto: f = m<br />
0<br />
+ m<br />
2<br />
+ m<br />
3<br />
+ m<br />
4<br />
+ m<br />
5<br />
+ m<br />
6<br />
+ m<br />
7<br />
2001. Febrero, primera semana (sistemas).<br />
✎ Simplifique la función lógica f(A,B,C,D)= M0 ⋅ M2 ⋅ M4 ⋅ M5 ⋅ M6 ⋅ M7 ⋅ M8 ⋅ M10 ⋅ M12⋅M13 ⋅ M14 ⋅ M15:<br />
Solución:<br />
f(A, B, C, D)<br />
=<br />
( A<br />
A B C D f(A, B, C, D)<br />
0 0 0 0<br />
0 0 0 1<br />
0 0 1 0<br />
0 0 1 1<br />
0 1 0 0<br />
0 1 0 1<br />
0 1 1 0<br />
0 1 1 1<br />
1 0 0 0<br />
1 0 0 1<br />
1 0 1 0<br />
1 0 1 1<br />
1 1 0 0<br />
1 1 0 1<br />
1 1 1 0<br />
1 1 1 1<br />
+ B + C + D)<br />
⋅ ( A + B + C + D)<br />
⋅ ( A + B + C + D)<br />
⋅ ( A + B + C + D) ⋅ ( A + B + C + D)<br />
⋅ ( A + B + C + D) ⋅<br />
(A + B + C + D)<br />
⋅ (A + B + C + D)<br />
⋅ (A + B + C + D)<br />
⋅ (A + B + C + D) ⋅ (A + B + C + D)<br />
⋅ (A + B + C + D)<br />
0<br />
0<br />
0<br />
0<br />
1<br />
0<br />
1<br />
0<br />
0<br />
0<br />
0<br />
0<br />
1<br />
0<br />
1<br />
0<br />
2001. Febrero, segunda semana (sistemas).<br />
CD<br />
A B<br />
C + D C + D C + D C + D<br />
A + B<br />
0 0<br />
A + B<br />
0 0 0 0<br />
A + B 0 0 0 0<br />
A + B<br />
0 0<br />
f(A, B, C, D) = B + D<br />
✎ Sea la función lógica <strong>de</strong> tres variables f(A,B,C)= A ⋅ B + C ⋅ A ⋅ B + B ⋅ C . Encuentre una forma canónica.<br />
Solución:<br />
A ⋅ B + C ⋅ A ⋅ B + B ⋅ C = A ⋅ B ⋅ C ⋅ A ⋅ B + B ⋅ C = (A + B)<br />
⋅ ( C + A + B) + B ⋅ C = (A + B + B ⋅ C) ⋅ ( C + A + B + B ⋅ C) =<br />
[ (A + B + B ) ⋅ (A + B + C) ] ⋅ [ ( C + A + B + B ) ⋅ (( C + A + B + C) ] = [ (A + B + B ) ⋅ (A + B + C) ] ⋅ [ 1⋅1]<br />
(A + B + 0 ) ⋅ (A + B + C) = (A + B + C ⋅ C) ⋅ (A + B + C) = (A + B + C ) ⋅ (A + B + C) ⋅ (A + B + C) = (A + B + C ) ⋅ (A + B + C) = M4<br />
⋅ M5<br />
2001. Febrero, segunda semana (Gestión).<br />
✎ Dada la función ( A + C + D)<br />
⋅ ( A + B + D)<br />
⋅ ( A + B + C)<br />
⋅ ( A + B + C)<br />
Solución:<br />
( A + C + D)<br />
= ( A + C + D)<br />
+ ( B ⋅ B)<br />
= ( A + C + D + B)<br />
⋅ ( A + C + D + B)<br />
= ( A + B + C + D)<br />
⋅ ( A + B + C + D)<br />
( A + B + D)<br />
= ( A + B + D)<br />
+ ( C ⋅ C)<br />
= ( A + B + D + C)<br />
⋅ ( A + B + D + C)<br />
= ( A + B + C + D)<br />
⋅ ( A + B + C + D)<br />
( A + B + C)<br />
= ( A + B + C)<br />
+ ( D ⋅ D)<br />
= ( A + B + C + D)<br />
⋅ ( A + B + C + D)<br />
( A + B + C)<br />
= ( A + B + C)<br />
+ ( D ⋅ D)<br />
= ( A + B + C + D)<br />
⋅ ( A + B + C + D)<br />
Por tanto:<br />
<strong>Ejercicios</strong> <strong>de</strong> <strong>Funciones</strong> <strong>Lógicas</strong> 12<br />
B<br />
D<br />
= (A + B ) ⋅ (A + B + C) =<br />
( A + C + D)<br />
⋅ ( A + B + D)<br />
⋅ ( A + B + C)<br />
⋅ ( A + B + C)<br />
=<br />
( A + B + C + D)<br />
⋅ ( A + B + C + D)<br />
⋅ ( A + B + C + D)<br />
⋅ ( A + B + C + D)<br />
⋅ ( A + B + C + D)<br />
⋅ ( A + B + C + D)<br />
⋅ ( A + B + C + D)<br />
⋅ ( A + B + C + D)<br />
M0<br />
⋅ M4<br />
⋅ M1<br />
⋅ M3<br />
⋅ M8<br />
⋅ M9<br />
⋅ M12<br />
⋅ M13<br />
= M0<br />
⋅ M1<br />
⋅ M3<br />
⋅ M4<br />
⋅ M8<br />
⋅ M9<br />
⋅ M12<br />
⋅ M13<br />
2001. Septiembre, original (sistemas).<br />
✎ Sea la función lógica <strong>de</strong> tres variables f(A,B,C)= ( A ⋅ B + C ⋅ A ⋅ B)<br />
⋅ ( B + C)<br />
. Encuentre una forma canónica.<br />
Solución:<br />
=
Change language