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+14 votes
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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 (96.2k points) | 1.4k views
0
$Remark:$ the complement of m0 is M0.
0
How?
0

Remember this formula- (A + B+ C) . (A + B + C') = (A + B)  remove that term which is not same.

5 Answers

+19 votes
Best answer
K-map
K-map

Answer is B

answered by Active (4.2k points)
edited 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

+10

ans: B

+5 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 (11.7k 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
q+q'=1

1+q'=1

qq'=0
0 votes
answer - B
answered by Loyal (8.7k 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 (4.5k points)
0 votes
Given Boolean Expression:

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

we know that dual of a boolean expression is equivalent. Take dual of it:

PQ'R'+PQ'R+PQR'

Now simplify it.(I am taking dual because solving the given expression will be difficult hence, take dual and solve it)

=PQ'(R'+R)+PQR'

=PQ'+PQR'

=P(Q'+QR')

=P(Q'+R').

Now if we see option b) and take dual of it then we will be getting the above simplified expression.
answered by (85 points)
Answer:

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