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The minterm expansion of $f(P,Q,R) = PQ +Q \bar{R}+P\bar{R}$ is

  1. $m_2+m_4+m_6+m_7$
  2. $m_0+m_1+m_3+m_5$
  3. $m_0+m_1+m_6+m_7$
  4. $m_2+m_3+m_4+m_5$
in Digital Logic
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0

order of function   P R Q weight of  P=4,Q=2,R=1)

absent variable we can take 0 and 1 both possibility 

total possible no   

PQ=(1110,111)=6,7

QR'=(010,110)=2,6

PR'=(100,110)=4,6

so min term (2,4,6,7)

option A

4 Answers

30 votes
 
Best answer

$PQ + QR' + PR' = PQR + PQR' + PQR' + P'QR' + PQR' + PQ'R' $

$\quad= PQR + PQR' + P'QR' + PQ'R' (111 + 110 + 010 + 100)$

$\quad =m_7 + m_6 + m_2 + m_4$

Option A.


Alternatively,
Using K-map


edited by
0
K-map is the best method to minimize the expression without any problem.
0
@K-map method is faster for me
0
but if take the counting of cell of k-map in vertical order the answer is comming;  m1+m4+m5+m7

and dont match any options. please explain
2

@raushan sah

You can expand  all the term

$f(P, Q, R)=PQ+Q\bar{R}+P\bar{R}$

$f(P, Q, R)=PQ(R+\bar{R})+(P+\bar{P})Q\bar{R}+P(Q+\bar{Q})\bar{R}$

$f(P, Q, R)=PQR+PQ\bar{R}+PQ\bar{R}+\bar{P}Q\bar{R}+PQ\bar{R}+P\bar{Q}\bar{R}$

$f(P, Q, R)=PQR+PQ\bar{R}+\bar{P}Q\bar{R}+P\bar{Q}\bar{R}$  [Remove the similar term]

In case of Sum of Product,we all know variable represent value $'1'$ and complement of variable represent the value $'0'.$

$f(P, Q, R)=PQR(111)+PQ\bar{R}(110)+\bar{P}Q\bar{R}(010)+P\bar{Q}\bar{R}(100)$

We can also represent,these minterm as

$f(P, Q, R)=\sum(7,6,4,2)$

and $m_{2}+m_{4}+m_{6}+m_{7}$ is also correct. [Because we write sum of min-term=sum of product term(Canonical)]

1
@lakshman  thanx

yes got it
0
nice @arjun sir
4 votes
option A

=PQ(R+~R)+(P+~P)Q~R+P~R(Q+~Q)

on simplification we will get minterms as 2,4,6,7
2 votes

Answer: OPTION A

–2 votes
answer -(A)
Answer:

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