in Digital Logic edited by
8,170 views
15 votes
15 votes

The simplified SOP (Sum of Product) from the Boolean expression

$$(\text{P} + \overline{\text{Q}} + \overline{\text{R}}) . (\text{P} + \text{Q + R) . (P + Q} +\overline{\text{R}})$$ is 

  1. $(\overline{\text{P}}.\text{Q}+\overline{\text{R}})$
  2. $(\text{P + Q}.\overline{\text{R}})$
  3. $(\text{P}.\overline{\text{Q}}+\text{R})$
  4. $\text{(P.Q + R)}$
in Digital Logic edited by
by
8.2k views

1 comment

i think B is right Answer
0
0

6 Answers

18 votes
18 votes
Best answer

$(P + \bar{Q} + \bar{R}) . (P + {Q} + R) . (P + Q +\bar{R})$

$(P + \bar{Q} + \bar{R}) . (P +(Q+R)(Q+\bar{R}))$

$(P + \bar{Q} + \bar{R}) . (P + Q)$

$(P + (\bar{Q}+ \bar{R})Q)$

$P+Q \bar{R}$

Hence,option(B)$P+Q \bar{R}$ is the correct choice.

edited by
13 votes
13 votes
ans is B

(p+q'+r')(p+q+r)(p+q+r')=(p+q'+r')(p+q)                    (since a=(a+b)(a+b') so putting p+q as a here)

=(p+q')(p+q) +(p+q)r' =p+pr'+qr'=p(1+r')+qr'=p+qr'     (again using above law and 1+anything=1
8 votes
8 votes

Answer is (B) part.

Another way to solve this problem is as follow -->

2 Comments

0,1 and 3 should be filled with 0 right?
1
1
yes
0,1 and 3 must be 0.
0
0
7 votes
7 votes
(P+Q'+R').(P+Q+R).(P+Q+R') is Standard POS which is eqt to ∐ (0,1,3) .

So Standard SOP =∑(2,4,5,6,7)  when we simplified by K-Map it will gives (P+QR').

Hence (B) (P+QR')  is Correct option.

1 comment

thank you ...
0
0