1.7k 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. $(P.Q+R)$
| 1.7k views
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$Remark:$ the complement of m0 is M0.
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How?
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Remember this formula- (A + B+ C) . (A + B + C') = (A + B)  remove that term which is not same.

by Active (4.2k points)
edited
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may be silly doubt  but m not getting this table , my POS terms = 4,5,6 but in kmap it shows 2,4,6

+12

ans: B

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if i solve by this method why i am getting wrong answer.

+1

@PRK

see the below k-map

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How could someone get this k map from these equations
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@Pardhyuman

In question, given expression is POS, so we can make K-map using them.

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

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

=P+PR+PQ+RP+RQ

=P(1+Q)+RP+RQ

=P(1+R)+RQ

=P+RQ
by Boss (12.1k points)
edited
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@puja Mishra how have you solved the question I did not get that plz elaborate
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See Boolean algebra rules ...
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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 :)
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Yo can also solve this using Kmap.
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depends on practice watever u prefer ....
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q+q'=1

1+q'=1

qq'=0
+1 vote
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.
by (105 points)
by Loyal (8.6k points)
(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'
by Active (4.8k points)