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+20 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

- $\bar{P}Q + Q\bar{R} + P\bar{Q}R$
- $\bar{P}Q + \bar{P}Q\bar{R} + PQ\bar{R} + P\bar{Q}R$
- $\bar{P}QR + \bar{P}Q\bar{R} + Q\bar{R} + P\bar{Q}R$
- $PQ\bar{R}$

+26 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)

\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)

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'))**

=

....solving further i got

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

+12

@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 ?.

Why we are taking S0 as MSB and S1 as LSB ?

We always take S0 as LSB, Does this because of representation in question ?.

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

@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

0

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

+8 votes

The Given inputs I_{1}(0), I_{2}(1), I_{3}(R), I_{4}(R)

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

Hence final Output F= (S0)'.(S1)'.I_{1} + (S0)'.(S1).I_{2 }+(S0).(S1)'.I_{3}+(S0).(S1).I_{4}

F= P'.Q'.I_{1} + P'.Q.I_{2} +P.Q'.I_{3} +P.Q.I_{4}

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

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