Question

A counter uses three D flip-flops generating the Gray code sequence 000, 001, 011, 010, 110, 111, 101, 100 repeatedly. The bits are in $Q_2 Q_1 Q_0$ format. The combinational logic expression for $D_1$ is \(\underline{\hspace{1cm}}\).

• A. $Q_2 Q_1 Q_0$
• B. $Q_2 Q_0 + Q_1 \overline{Q_0}$
• C. $\overline{Q_2} Q_0 + Q_1 \overline{Q_0}$
• D. $Q_2 Q_1 + \overline{Q_2} \overline{Q_1}$

Hint:

When designing counters, always map present states to next states and extract bitwise transitions to derive the D-input logic.

Answer 1

The Correct Option is C

The Gray-code sequence is:
000 → 001 → 011 → 010 → 110 → 111 → 101 → 100 → repeat.
We extract the present state $(Q_2, Q_1, Q_0)$ and the next value of $Q_1$ (which equals $D_1$):
\[ \begin{array}{c|c} (Q_2 Q_1 Q_0)_{\text{present}} & Q_1^{+} = D_1
\hline 000 & 0
001 & 1
011 & 1
010 & 1
110 & 1
111 & 0
101 & 0
100 & 0
\end{array} \] Now write minterms where $D_1 = 1$:
States giving $D_1 = 1$ are: 001, 011, 010, 110.
Corresponding minterms:
001 → $\overline{Q_2}\,\overline{Q_1}\,Q_0$
011 → $\overline{Q_2}\,Q_1\,Q_0$
010 → $\overline{Q_2}\,Q_1\,\overline{Q_0}$
110 → $Q_2\,Q_1\,\overline{Q_0}$
Group and simplify using K-map:
Resulting simplified expression is: \[ D_1 = \overline{Q_2}\,Q_0 \;+\; Q_1\,\overline{Q_0}. \] This matches option (C).
Final Answer: $\overline{Q_2} Q_0 + Q_1 \overline{Q_0}$