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+12 votes
1.1k views

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

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

  1. $(\bar{P}.Q+\bar{R})$
  2. $(P+\bar{Q}.\bar{R})$
  3. $(\bar{P}.Q+R)$
  4. $(P.Q+R)$
asked in Digital Logic by Veteran (106k points) | 1.1k views
0
$Remark:$ the complement of m0 is M0.
0
How?

4 Answers

+19 votes
Best answer
Karnaugh map
1 0 0 0
1 1 1

1

Answer is B

answered by Active (4.2k points)
selected by
0

may be silly doubt  but m not getting this table , my POS terms = 4,5,6 but in kmap it shows 2,4,6

joshi_nitish

+6

ans: B

+3 votes
(P+Q'+R').(P+Q'+R).(P+Q+R')

=(P+R`)(Q+Q`)(P+Q`+R)

=P+PR+PQ`+R`P+R`Q`

=P(1+Q`)+R`P+R`Q`

=P(1+R`)+R`Q`

=P+R`Q`
answered by Boss (11.6k points)
edited by
0
@puja Mishra how have you solved the question I did not get that plz elaborate
0
See Boolean algebra rules ...
0
what rule did you apply for : (P+Q¯+R¯).(P+Q¯+R).(P+Q+R¯)=(P+R`)(Q+Q`)(P+Q`+R)
+1
(A+B)(A+B')= A
+1
Thank you :)
0
Yo can also solve this using Kmap.
0
depends on practice watever u prefer ....
0 votes
answer - B
answered by Loyal (9.1k points)
0 votes
(P+Q'+R') (P+Q'+R) (P+Q+R')

complement the whole equation

P'QR+P'QR'+P'Q'R

taking P'R common from 1st and 3rd minterm

P'R(Q+Q') + P'QR'

we know A+A' = 1

P'R+P'QR'

P'(R+QR')

P'((R+Q)(R+R'))

P'(R+Q)

Complement it back

P+R'Q'

P+Q'R'
answered by Active (1.5k points)
Answer:

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