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Official PYQ · 2014Set 1 · Q45Combinational CircuitsMCQ1 mark

GATE CS 2014 Set 1 Q45 — Combinational Circuits

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Consider the 4-to-14\text{-to-1} multiplexer with two select lines S1S_1 and S0S_0 given below

The minimal sum-of-products form of the Boolean expression for the output FF of the multiplexer is

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Answer: A

The output FF of a 4-to-14\text{-to-}1 multiplexer with select lines S1,S0S_1, S_0 and inputs I0,I1,I2,I3I_0, I_1, I_2, I_3 is given by:

F=S1ˉS0ˉI0+S1ˉS0I1+S1S0ˉI2+S1S0I3F = \bar{S_1}\bar{S_0}I_0 + \bar{S_1}S_0I_1 + S_1\bar{S_0}I_2 + S_1S_0I_3

From the given circuit diagram, we have:

  • Select lines: S1=PS_1 = P and S0=QS_0 = Q
  • Inputs: I0=0I_0 = 0, I1=1I_1 = 1, I2=RI_2 = R, and I3=RˉI_3 = \bar{R}

Substituting these values into the output equation:

F=PˉQˉ(0)+PˉQ(1)+PQˉ(R)+PQ(Rˉ)F = \bar{P}\bar{Q}(0) + \bar{P}Q(1) + P\bar{Q}(R) + PQ(\bar{R}) F=PˉQ+PQˉR+PQRˉF = \bar{P}Q + P\bar{Q}R + PQ\bar{R}

To find the minimal sum-of-products (SOP) form, we simplify the expression using the absorption law (A+AˉB=A+BA + \bar{A}B = A + B):

F=PˉQ+PQRˉ+PQˉR=Q(Pˉ+PRˉ)+PQˉR=Q(Pˉ+Rˉ)+PQˉR=PˉQ+QRˉ+PQˉR\begin{aligned} F &= \bar{P}Q + PQ\bar{R} + P\bar{Q}R \\ &= Q(\bar{P} + P\bar{R}) + P\bar{Q}R \\ &= Q(\bar{P} + \bar{R}) + P\bar{Q}R \\ &= \bar{P}Q + Q\bar{R} + P\bar{Q}R \end{aligned}

This matches Option A.

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