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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. $PQ\bar{R}$
edited | 2.2k views
+10

this image is wrong and Answer is option A

0
image is correct

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'

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)

selected by
0

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?

+9
@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 ?.
+2
@Sachin I also have same confusion. Please explain @Arjun Sir
+1
Folks I also have same confusion. @Arjun Sir please explain this concept if possible.
0
Actually option matching is main criteria here

For other options answer not matching
0
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
+1
@srestha

since option doesnt match okay

otherwise S0 is always the LSB right
0
yes, that should be
0
A+A′B=A+B - which law or rule in logic design is this?
+1

@sachinkodagali You can easily derive it,

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

= (1) (A+B)

=(A+B)

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

+1
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
+2

@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

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

0

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)

¯Q+P¯QR¯+PQR¯+PQ¯R

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
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)

1
2