Q′Q'(P⊕R)+Q(P⊕R')′

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+10 votes

The Boolean expression of the output $f$ of the multiplexer shown below is

- $\overline {P \oplus Q \oplus R}$
- $P \oplus Q \oplus R$
- $P+Q+R$
- $\overline{P+Q+R}$

+24 votes

Best answer

$f = S_0'S_1' R + S_0'S_1R' + S_0S_1'R' + S_0S_1R$

$\quad=Q'P'R + Q'PR' + QP'R' + QPR $

$\quad= Q'(P⊕R) + Q(P⊕R)' $

$\quad= Q⊕P⊕R = P⊕Q⊕R$

Doing truth value substitution,

$${\begin{array}{|cccc|c|}\hline

\textbf{P}& \textbf{Q}& \textbf{R}&\bf{f}& \bf{P \oplus Q \oplus R } \\\hline

0&0&0&0&0 \\ 0&0&1&1&1\\ 0&1&0&1&1\\ 0&1&1&0&0\\ 1&0&0&1&1\\ 1&0&1&0& 0 \\ 1&1&0&0&0\\ 1&1&1&1& 1\\ \hline

\end{array}}$$

$\quad=Q'P'R + Q'PR' + QP'R' + QPR $

$\quad= Q'(P⊕R) + Q(P⊕R)' $

$\quad= Q⊕P⊕R = P⊕Q⊕R$

Doing truth value substitution,

$${\begin{array}{|cccc|c|}\hline

\textbf{P}& \textbf{Q}& \textbf{R}&\bf{f}& \bf{P \oplus Q \oplus R } \\\hline

0&0&0&0&0 \\ 0&0&1&1&1\\ 0&1&0&1&1\\ 0&1&1&0&0\\ 1&0&0&1&1\\ 1&0&1&0& 0 \\ 1&1&0&0&0\\ 1&1&1&1& 1\\ \hline

\end{array}}$$

+1

@Arjun sir, dosent the answer hold good for option A too?? as we have odd number of variables , xor and xnor give same ans right ?

0

@Arjun sir, dosent the answer hold good for option A too?? as we have odd number of variables , xor and xnor give same ans right ?

@Arjun sir same doubt. :(

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