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Consider the following Boolean expression for F:

$F(P,Q,R,S)= PQ + \bar{P}QR + \bar{P}Q\bar{R}S$

The minimal sum$-$of$-$products form of $F$ is

1. $PQ+QR+QS$
2. $P+Q+R+S$
3. $\bar{P} + \bar{Q}+ \bar{R}+ \bar{S}$
4. $\bar{P}R + \bar{R} \bar{P}S+P$

 $PQ\ RS$ $\bar R \bar S$ $\bar R S$ $RS$ $R \bar S$ $\bar P \bar Q$ 0 0 0 0 $\bar P Q$ 0 1 1 1 $PQ$ 1 1 1 1 $P \bar Q$ 0 0 0 0

Minimal SOP $= PQ + QR + QS$

Hence, option A is correct.

selected

so Option A is correct

Solution:

F= PQ+P’QR+P’QR’S =Q(P+P’+P’R’S)

USE PROP a+a’b=a+b =Q(P+R+P’R’S) =Q(P+P’R’S+R) USE PROP a+a’b=a+b =Q(P+R’S+R) =Q(P+R+R’S) USE PROP a+a’b=a+b =Q(P+R+S) =PQ+QR+QS Ans (a)
ans a)
option A
–1 vote

Ans. is A

–1 vote

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