The GATE Grind

GATE 2018 EC – Question 41

Digital Circuits · Combinatorial Circuits · 2 marks · Multiple choice

A four-variable Boolean function is realized using $4\times1$ multiplexers as shown in the figure.

The minimized expression for $F(U,V,W,X)$ is

Diagram for GATE 2018 EC question 41
  1. $(UV+\bar U\bar V)\bar W$
  2. $(UV+\bar U\bar V)(\bar W\bar X+\bar WX)$
  3. $(U\bar V+\bar UV)\bar W$
  4. $(U\bar V+\bar UV)(\bar W\bar X+\bar WX)$

Practise this question in The GATE Grind →

Show answer and explanation

Correct answer: (C) $(U\bar V+\bar UV)\bar W$

Explanation

For the first multiplexer the inputs are $I_0=0$, $I_1=I_2=1$ and $I_3=0$, so its output is $M=\bar UV+U\bar V$. For the second, $I_0=I_1=M$ and $I_2=I_3=0$, so the output is $M$ when $W=0$ and 0 when $W=1$. Hence $F=(U\bar V+\bar UV)\bar W$.