Question 1.25: Draw the circuit for Y=A·(B+C) and produce alternative repre...

The original circuit, and the effects of replacing the NOR and AND operators, respectively, are shown in Fig. 1.17. So:

which can be implemented using three-input AND and OR gates respectively.

Using Boolean algebra, since (\overline{B+C})=(\overline{B}·\overline{C}) then Y=A·\overline{B} ·\overline{C}, the AND gate implementation. To obtain the sum expression (OR gate implementation):

\overline{Y} =\overline{A·(\overline{B+C} )}    inverting both sides

     =\overline{A}+ (\overline{\overline{B+C}})    de Morgan’s theorem

Hence Y=\overline{\overline{A} +B+C} as above.

The truth table for these is given in Table 1.6. Note that the AND gate produces a 1 when A = 1 and B= C=0, which specifies a single row in the truth table. The OR operator can be considered in a slightly different way. This is that Y=0 (or alternatively \overline{Y} =1) when either (A=0) OR (B= 1) OR (C= 1). This is again a consequence of duality.

Table 1.6 Truth table relating to Example 1.25