GATE CS 2013 Q5 — Combinational Circuits

The given truth table represents a Priority Encoder (specifically, a 4-to-2 priority encoder).

Analysis of the Truth Table:

• Inputs: $D_0, D_1, D_2, D_3$
• Outputs: $X_1, X_0$ (binary representation of the active input index) and $V$ (valid bit indicator).

Let us observe the priority behavior from the rows of the table:

1. If $D_3 = 1$, the output is $X_1 X_0 = 11$ (binary for 3) regardless of the states of $D_2, D_1, D_0$ (indicated by don't cares 'x'). This shows $D_3$ has the highest priority.
2. If $D_3 = 0$ and $D_2 = 1$, the output is $X_1 X_0 = 10$ (binary for 2) regardless of $D_1, D_0$.
3. If $D_3 = 0, D_2 = 0$ and $D_1 = 1$, the output is $X_1 X_0 = 01$ (binary for 1) regardless of $D_0$.
4. If $D_3 = 0, D_2 = 0, D_1 = 0$ and $D_0 = 1$, the output is $X_1 X_0 = 00$ (binary for 0).
5. If all inputs are $0$ ($0000$), the valid bit $V = 0$, indicating that no input is active. For all other cases, $V = 1$.

Since the circuit encodes the index of the active input line with the highest priority into a binary code, it is a Priority Encoder.