Question

In a given sequential circuit, initial states are $Q_1 = 1$ and $Q_2 = 0$. For a clock frequency of 1 MHz, the frequency of signal $Q_2$ in kHz, is ___________ (rounded off to the nearest integer). 

 

Hint:

For any synchronous sequential circuit, the best way to determine the output frequency is to create a state table and trace the sequence of states and outputs. The number of states in the repeating cycle determines the period of the output signal in terms of clock cycles.

Answer 1

Correct Answer: 250

To solve this problem, we need to determine the frequency of the signal $Q_2$. The circuit comprises two D flip-flops connected sequentially.
Given the initial states: $Q_1 = 1$, $Q_2 = 0$ and a clock frequency of 1 MHz.

Steps to determine the frequency of $Q_2$:

1. The D flip-flop transfers its input to its output at the clock's rising edge. Thus, for the first flip-flop, if $D_1 = Q_{\text{prev}}$ (previous clock cycle's value), then $Q_1 = D_1$.
2. In the second flip-flop, $Q_2 = D_2$ and $D_2$ is the output of the first flip-flop, $Q_1$. Therefore, $Q_2 = Q_1$ of the previous cycle.
3. Considering the clock frequency of 1 MHz, a complete cycle (both rising and falling edge) will halve the frequency for $Q_1$, making it 500 kHz.
4. Since $Q_2$ also follows $Q_1$ with a delay of one clock cycle, $Q_2$ gets halved again, resulting in:

Frequency of $Q_2 = \frac{500\, \text{kHz}}{2} = 250\, \text{kHz}$.

The calculated frequency is 250 kHz, which is within the given range of [250,250]. Therefore, the frequency of signal $Q_2$ is indeed 250 kHz.