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+16 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. $PQ\bar{R}$
asked in Digital Logic by Veteran (115k points)
edited by | 2.7k views

this image is wrong and Answer is option A

image is correct

I seriously don't know which image is correct.
Source of GATE paper from IITR displays a different image.

Here S1=P, and S0=Q

4 Answers

+25 votes
Best answer

S0 and S1 are used to select the input given to be given as output.

S0 S1 Output
0 0 0
0 1 1
1 0 R
1 1 R'


\bf{s_0}&    \bf{s_1}&  \textbf{Output} \\\hline
0&0&0 \\\hline0&1&1 \\ \hline   1&0&\text{R} \\ \hline   1&1&\text{R'}\\ \hline  

So, output becomes 1 for
$S0'S1 + S0S1'R + S0S1R'$

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

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

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

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

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

Option (A)

answered by Veteran (396k 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 

+6 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


answered by (403 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)


+3 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
answered by Boss (42.7k 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)
answered by Loyal (9.3k points)

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