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+10 votes
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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. $\overline{P+Q+R}$
asked in Digital Logic by Veteran (115k points)
edited by | 2.2k views

2 Answers

+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,
$${\begin{array}{|cccc|c|}\hline
\textbf{P}&    \textbf{Q}&  \textbf{R}&\bf{f}& \bf{P \oplus Q \oplus R } \\\hline
0&0&0&0&0 \\ 0&0&1&1&1\\     0&1&0&1&1\\    0&1&1&0&0\\    1&0&0&1&1\\  1&0&1&0& 0  \\    1&1&0&0&0\\    1&1&1&1& 1\\ \hline
\end{array}}$$
answered by Veteran (396k points)
edited by
+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 ?
0

Let : Y = P XOR R

So : Q' Y + Q Y' 

= Q XOR Y

= Q XOR P XOR R

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. :(

0 votes
answer - B
answered by Loyal (8.8k points)
Answer:

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