Sequential Circuits

From the circuit, \(D_1 = Q_0\), \(D_2 = Q_1\), \(D_3 = Q_2\) and \(D_0 = \overline{Q_3}\). Each clock shifts the bits one place and feeds the complement of the last bit back to the first: a 4-bit Johnson (twisted-ring) counter.

With the state written as \(Q_0Q_1Q_2Q_3\): 0000 → 1000 → 1100 → 1110 → 1111 → 0111 → 0011 → 0001 → 0000. The other 8 states form a second cycle of the same length (0100 → 1010 → 1101 → 0110 → 1011 → 0101 → 0010 → 1001 → 0100), so from any initial state the circuit comes back after 8 clocks.

That is 8 distinct states counting the initial one, or 7 besides it, so the official key accepts 7 to 8.

Answer: 8 (7 is also accepted)

Each flip-flop in a ripple counter toggles on every active edge it receives, so its output completes one cycle for every two input cycles: each stage halves the frequency. The output of the 4th (last) flip-flop therefore runs at \(f_{clk}/2^{4} = f_{clk}/16\), and its period is 16 clock periods.

\(16\,T_{clk} = 64\ \mu\text{s} \Rightarrow T_{clk} = 4\ \mu\text{s} \Rightarrow f_{clk} = \frac{1}{4\ \mu\text{s}} = 250\ \text{kHz}\).

The counter’s frequency is its input clock frequency. Taking the last stage’s own frequency, \(1/64\ \mu\text{s} \approx 15.6\) kHz, is the usual slip.

Answer: 250