PUERTA | PUERTA CS 2010 | Pregunta 65 – Part 6

¿Cuál es la expresión booleana para la salida f del circuito lógico combinacional de las puertas NOR que se muestra a continuación?

(A) (Q+R)’
(B) (P+Q)’
(C) (P+R)
(D) (P+Q+R)’

Respuesta: (A)
Explicación: La respuesta es la opción A.

La pregunta anterior contiene las puertas NOR. Veamos qué hace la puerta NOR.
Si A y B son las dos entradas de la puerta NOR, la puerta NOR da (A+B)’ como salida.

Asignemos números a las Puertas para facilitar la comprensión.

In the 1st column there are 4 NOR Gates, 
number them as 1 to 4 ( top to down).

In the 2nd column there are 2 NOR Gates, 
number them as 5 and 6 ( top to down).

In the 3rd column there is only 1 NOR Gate, 
number it as 7.

1st numbered Gate gives output as : ( P + Q )'
2nd numbered Gate gives output as : ( Q + R )'
3rd numbered Gate gives output as : ( P + R )'
4th numbered Gate gives output as : ( R + Q )'
5th numbered Gate gives output as :
(( P + Q )' + ( Q + R )')'
= ((P + Q)'' . ( Q + R )'') ( De Morgan's law)
= (P + Q ) . ( Q + R ) ( Idempotent Law, A'' = A)
= (PQ + PR + Q + QR )
= (Q(1 + P + R) + PR) = Q + PR ( as, 1 + " any 
                                boolean expression" = 1 )

Similarly 6th numbered Gate gives output as : R + PQ 
                             (as this time R is common here)

Now 7th numbered Gate gives output as :
((Q + PR) + (R + PQ))'
= (Q( 1+P) + R(1+P))'
= (Q+R)' 

Cuestionario de esta pregunta