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+22 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. $PQ\bar{R}$
in Digital Logic by Veteran (106k points)
edited by | 3.6k views
If we simplify option (C) then also we get P'Q+QR'+PQ'R then  how option a) and c) are different ?

4 Answers

+27 votes
Best answer
$S_0$ 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)
by Veteran (434k points)
edited by

I understood the approach, but can someone please explain what is wrong in below approach...
P'Q + PQ'R + PQR'
= P'Q + P(Q'R + QR')
= (P'Q + P) (P'Q + (Q'R +QR'))
= (P + P') (P + Q) (P'Q + (Q'R +QR'))

(P + Q) (P'Q + Q'R +QR')
....solving further i got
: P + QR' + P'Q. 

is there something wrong in my approach or its just an answer which  is not given in options?

@Arjun sir, plz help me with one confusion.

Why we are taking S0 as MSB and S1 as LSB ?
We always take S0 as LSB, Does this because of representation in question ?.
@Sachin I also have same confusion. Please explain @Arjun Sir
Folks I also have same confusion. @Arjun Sir please explain this concept if possible.
Actually option matching is main criteria here

For other options answer not matching
i think we should consider whatever is given on the right most end always as most significant bit as the charecteristic equation is written according to that considering Q is connected to the LSB and P is connected to the MSB because the circuit inside is wired in that manner or else we can go by options

since option doesnt match okay

otherwise S0 is always the LSB right
yes, that should be
A+A′B=A+B - which law or rule in logic design is this?

@sachinkodagali You can easily derive it,

A+A'B= (A+A') (A+B)   : as + distributes over dot

          = (1) (A+B)


P.S.: Don't remember these simple things just derive it in the exam hall, It won't take much time.

Can anyone help me why we are taking S0 as MSB instead of taking it as LSB (leaving behind checking from options) @Arjun Suresh Sir please help here

@Sandeep Suri have a look at the multiplexers circuit diagram u will understand that the variable written in the leftmost side let it be S0 OR S1 is always connected in the circuitry inside in such a way that u get the respective minterms as output function there is no need to follow convention always based on what is connected to inside circuit u take the notation 


@Arjun sir according to the question $S_{0} = Q$ and $S_{1} = P$. But you took $S_{0}$ as $P$ and $S_{1}$ as $Q$.


@Arjun sir please correct the diagram.becuase in GO pdf diagram is  bit different.

+9 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


by (423 points)

F=P'Q + RPQ+ R'PQ = Q( P'+R'P) + R'PQ = Q(P'+ R')  +R'PQ

P' +R'P = P'+R'  SO ANSWER IS (A)


+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
by Boss (44.3k points)
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)
by Boss (10k points)

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