GATE 2022 EC Q.19: solution and answer

GATE 2022 · ECQ.19 · 1 mark · MCQ

Consider the 2-bit multiplexer (MUX) shown in the figure. For OUTPUT to be the XOR of C and D, the values for $A_{0}$ , $A_{1}$ , $A_{2}$ , and $A_{3}$ are ________.

GATE 2022 EC question 19: A 4-to-1 multiplexer drawn as a box labelled with data inputs 0, 1, 2, 3 on its left side. — Figure from the GATE 2022 EC paper. It is part of the question.

(A)
$A_{0} = 0, A_{1} = 0, A_{2} = 1, A_{3} = 1$
(B)
$A_{0} = 1, A_{1} = 0, A_{2} = 1, A_{3} = 0$
(C)
$A_{0} = 0, A_{1} = 1, A_{2} = 1, A_{3} = 0$
(D)
$A_{0} = 1, A_{1} = 1, A_{2} = 0, A_{3} = 0$

Answer and solution

ANSWER

$A_{0} = 0, A_{1} = 1, A_{2} = 1, A_{3} = 0$ , which is option (C).

Official answer: CSource: Official answer key, GATE 2022 EC (IIT Kharagpur)

How a 4-to-1 MUX works: the output copies the data input whose index is the binary number on the select lines, $S_{1} S_{0}$ . Here $S_{1} = C$ (the MSB) and $S_{0} = D$ (the LSB), so $OUTPUT = A_{(C D)_{2}}$ .
Write out the output: $OUTPUT = \overline{C} \overline{D} A_{0} + \overline{C} D A_{1} + C \overline{D} A_{2} + C D A_{3}$ (an overline means complement).
The function we want: $C \oplus D = \overline{C} D + C \overline{D}$ , which is 1 only for minterms $m_{1}$ ( $C D = 01$ ) and $m_{2}$ ( $C D = 10$ ).
Match the two expressions minterm by minterm: each data input $A_{i}$ appears only in minterm $m_{i}$ , so it must equal the function's value there: $A_{i} = f (m_{i})$ .
$C D = 00 \Rightarrow$ input 0 is selected, and $0 \oplus 0 = 0$ , so $A_{0} = 0$ .
$C D = 01 \Rightarrow$ input 1 is selected, and $0 \oplus 1 = 1$ , so $A_{1} = 1$ .
$C D = 10 \Rightarrow$ input 2 is selected, and $1 \oplus 0 = 1$ , so $A_{2} = 1$ .
$C D = 11 \Rightarrow$ input 3 is selected, and $1 \oplus 1 = 0$ , so $A_{3} = 0$ .
So $(A_{0}, A_{1}, A_{2}, A_{3}) = (0, 1, 1, 0)$ , which is option (C). Check: substituting into step 2 gives $OUTPUT = \overline{C} D + C \overline{D} = C \oplus D$ .

KEY IDEA

When the function's variables drive a MUX's select lines, the data inputs are simply the function's truth-table column: $A_{i} = f (m_{i})$ , with $m_{i}$ indexed by $S_{1} S_{0}$ . For XOR that column is $0, 1, 1, 0$ . XOR is symmetric in C and D, so the answer does not change even if you read $S_{0}$ as the MSB, because swapping the select lines only swaps $A_{1}$ and $A_{2}$ , and both are 1.

WATCH OUT FOR

Picking (A) (0, 0, 1, 1), which is just $C$ (the MSB select line), or (B) (1, 0, 1, 0), which is $\overline{D}$ . This happens when the truth table is written against one variable instead of against the combined index $(S_{1} S_{0})_{2}$ . Picking (D) means confusing XOR with its complement or with $\overline{C}$ .

From the GATE 2022 EC paper. Every question of this paper · Download the full paper · All 12 Combinational questions, with notes

GATE 2022 EC Q.19: solution and answer

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