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37 votes
37 votes

Consider the $4\text{-to-1}$ multiplexer with two select lines $ S_1$ and $ S_0 $ given below

 

The minimal sum-of-products form of the Boolean expression for the output $F$ of the multiplexer is

  1. $\bar{P}Q + Q\bar{R} + P\bar{Q}R$
  2. $\bar{P}Q + \bar{P}Q\bar{R} + PQ\bar{R} + P\bar{Q}R$
  3. $\bar{P}QR + \bar{P}Q\bar{R} + Q\bar{R} + P\bar{Q}R$
  4. $PQ\bar{R}$
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4 Answers

Best answer
36 votes
36 votes
$S_0$ and $S_1$ are used to select the input given to be given as output.$${\begin{array}{|c|c|c|}\hline
\bf{S_0}&    \bf{S_1}&  \textbf{Output} \\\hline
0&0&0 \\0&1&1 \\    1&0&\text{R} \\   1&1&\text{R'}\\ \hline
\end{array}}$$So, output becomes $1$ for
$S_0'S_1 + S_0S_1'R + S_0S_1R'$

$=P'Q + PQ'R + PQR'$

$=P'Q + PQR' + PQ'R$

$=Q(P'+PR') + PQ'R$

$=Q(P' + R') + PQ'R \text{ } (\because A+A'B = A+B)$
$ = P'Q + QR' + PQ'R$

Option (A)
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14 votes
14 votes

The Given inputs I1(0),  I2(1), I3(R), I4(R)

and Selection line S0(P) and S1(Q)

Hence final Output  F= (S0)'.(S1)'.I1 + (S0)'.(S1).I2   +(S0).(S1)'.I3+(S0).(S1).I4 

                               F= P'.Q'.I1 + P'.Q.I2 +P.Q'.I3 +P.Q.I4

                              F=P'.Q'.0 + P'.Q.1+P.Q'.R +P.Q.R'

                              F= P'.Q.+P.Q'.R +P.Q.R'  but this equation does not match any option so we draw K-Map

                         

4 votes
4 votes
Initial SOP expression comes to P'Q + PQ'R + PQR'   but we don't have any such option, make a k-map and minimize it, and option A will match with the answer
0 votes
0 votes
For 4 to 1 mux =p’q’(0)+p’q(1)+pq’r+pqr’ =p’q+pq’r+pqr’ =q(p’+pr’)+pq’r =q(p’+r’)+pq’r =p’q+qr’+pq’r

Ans (a)
Answer:

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