2.7k views

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

1. $\overline {P \oplus Q \oplus R}$
2. $P \oplus Q \oplus R$
3. $P+Q+R$
4. $\overline{P+Q+R}$

edited | 2.7k views

$f = S_0'S_1' R + S_0'S_1R' + S_0S_1'R' + S_0S_1R$
$\quad=Q'P'R + Q'PR' + QP'R' + QPR$
$\quad= Q'(P⊕R) + Q(P⊕R)'$
$\quad= Q⊕P⊕R = P⊕Q⊕R$

Doing truth value substitution,
$${\begin{array}{|cccc|c|}\hline \textbf{P}& \textbf{Q}& \textbf{R}&\bf{f}& \bf{P \oplus Q \oplus R } \\\hline 0&0&0&0&0 \\ 0&0&1&1&1\\ 0&1&0&1&1\\ 0&1&1&0&0\\ 1&0&0&1&1\\ 1&0&1&0& 0 \\ 1&1&0&0&0\\ 1&1&1&1& 1\\ \hline \end{array}}$$

Correct Answer: $B$
by Veteran (431k points)
edited
+1

QQ'(PR)+Q(PR')
HOW TO SIMPLIFY IT:
+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 ?
+2

Let : Y = P XOR R

So : Q' Y + Q Y'

= Q XOR Y

= Q XOR P XOR R

+2

@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. :(

0

@tusharp assume option(b) as ${X}$ then option(a) becomes ${X}'$, how can they both be equal... option(a) would have been true if (Xnor) was used in between the symbols.

by Loyal (8.6k points)