GATE CS 2013 Q.5 (CS Q.5): Combinational Circuits

In the following truth table, $V = 1$ if and only if the input is valid.

What function does the truth table represent?

D0	D1	D2	D3	X0	X1	V
0	0	0	0	x	x	0
1	0	0	0	0	0	1
x	1	0	0	0	1	1
x	x	1	0	1	0	1
x	x	x	1	1	1	1

Priority encoder

Decoder

Multiplexer

Demultiplexer

Official answer and solution

Official answer: A
From: Official GATE 2013 answer key

Solution

Analyze the input-output mapping in the truth table and identify the digital logic circuit it specifies.

Examine the truth table:

When all inputs $D_0, D_1, D_2, D_3$ are 0, the valid output $V = 0$ (no valid active input).
When $D_3 = 1$ (highest priority), the output is $(X_1, X_0) = (1, 1)$ , regardless of $D_2, D_1, D_0$ .
When $D_3 = 0$ and $D_2 = 1$ , the output is $(X_1, X_0) = (1, 0)$ , regardless of $D_1, D_0$ .
When $D_3 = D_2 = 0$ and $D_1 = 1$ , the output is $(X_1, X_0) = (0, 1)$ , regardless of $D_0$ .
When $D_3 = D_2 = D_1 = 0$ and $D_0 = 1$ , the output is $(X_1, X_0) = (0, 0)$ .

This behavior, where the output code reflects the highest-indexed active input line and don't-care conditions apply to lower lines, precisely defines a 4-to-2 Priority Encoder with a valid bit $V$ .

Final answer

Option (A)

Common trap

Confusing a priority encoder (multiple inputs mapped to binary code based on priority) with a standard decoder or multiplexer.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2013 Computer Science and Information Technology (CS)
Question: 5 of 65, printed as CS Q.5
Section: Computer Science and Information Technology

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