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GATE 2016 CS – Question 40

Digital Logic · Combinational Circuits · 2 marks · Multiple choice

Consider the two cascaded 2-to-1 multiplexers as shown in the figure.

[Figure: The first 2-to-1 MUX has input 0 tied to 0, input 1 tied to $R$, and select $P$. Its output goes to input 1 of the second 2-to-1 MUX. The second MUX has input 0 tied to $\bar{R}$ and select $Q$, and its output is $X$.]

The minimal sum of products form of the output $X$ is

Diagram for GATE 2016 CS question 40
  1. $\bar{P}\bar{Q} + PQR$
  2. $\bar{P}Q + QR$
  3. $PQ + \bar{P}\bar{Q}R$
  4. $\bar{Q}\bar{R} + PQR$

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Correct answer: (D) $\bar{Q}\bar{R} + PQR$

Explanation

The first multiplexer outputs $\bar{P} \cdot 0 + P \cdot R = PR$. The second multiplexer selects $\bar{R}$ when $Q = 0$ and $PR$ when $Q = 1$, so $X = \bar{Q}\bar{R} + Q \cdot PR = \bar{Q}\bar{R} + PQR$.