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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. $\bar{P}R + \bar{R} \bar{P}S+P$
asked in Digital Logic by Veteran (100k points) | 733 views

8 Answers

+9 votes
Best answer

 

$PQ\ RS$ $\bar R \bar S$ $\bar R S$ $RS$ $R \bar S$
$\bar P \bar Q$ 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.
 

 

answered by Veteran (29.9k points)
selected by
+4 votes

so Option A is correct

answered by Active (1.8k points)
+1 vote
ans a)
answered by Boss (5.2k points)
+1 vote
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)
answered by Boss (5.9k points)
0 votes
option A
answered by Loyal (2.9k points)
0 votes

Ans. is A

answered by (391 points)
0 votes
A. ANSWER IS CORRECT..
answered by (11 points)
0 votes

Solution:

 

answered ago by Active (2.1k points)


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