Question

The clock frequency of the digital circuit shown in the figure is 12 MHz. The frequencies of the output (F) corresponding to Control = 0 and Control = 1, respectively, are

 

• A. 4 MHz and 6 MHz
• B. 6 MHz and 4 MHz
• C. 3 MHz and 4 MHz
• D. 3 MHz and 6 MHz

Hint:

Carefully analyze the logic expressions for the D flip-flop inputs based on the control signal and the current states. Trace the state transitions over multiple clock cycles to determine the frequency of the output.

Answer 1

The Correct Option is A

Let the state of the first flip-flop be \( Q_1 \) and the second be \( Q_2 \) (which is the output \( F \)). The inputs to the flip-flops are:
\( D_1 = Q_1 \overline{{Control}} + \overline{Q_1} {Control} = Q_1 \oplus {Control} \)
\( D_2 = Q_1 {Control} + \overline{Q_1} \overline{{Control}} = Q_1 \odot {Control} = \overline{Q_1 \oplus {Control}} \)
Case 1: Control = 0
\( D_1 = Q_1 \oplus 0 = Q_1 \)
\( D_2 = \overline{Q_1 \oplus 0} = \overline{Q_1} \)
The first flip-flop holds its state. The second flip-flop takes the complement of the first. After the first clock cycle, both \( Q_1 \) and \( Q_2 \) become stable (either 0 or 1), so the frequency of the output \( F = Q_2 \) is 0 Hz in steady state. This does not match any of the options.

Case 2: Control = 1
\( D_1 = Q_1 \oplus 1 = \overline{Q_1} \)
\( D_2 = \overline{Q_1 \oplus 1} = \overline{\overline{Q_1}} = Q_1 \)
The state transitions are:
\((Q_1, Q_2) \rightarrow (\overline{Q_1}, Q_1) \rightarrow (\overline{\overline{Q_1}}, \overline{Q_1}) = (Q_1, \overline{Q_1}) \rightarrow (\overline{Q_1}, Q_1) \rightarrow \dots\)
The sequence of \( Q_1 \) is \( Q_1, \overline{Q_1}, Q_1, \overline{Q_1}, \dots \) with a frequency of \( \frac{12}{2} = 6 \) MHz.
The sequence of \( Q_2 \) (output \( F \)) is \( Q_2(0), Q_1(0), \overline{Q_1(0)}, Q_1(0), \dots \)
If \( Q_1(0) = 0 \), \( F \): ?, 0, 1, 0, ...
If \( Q_1(0) = 1 \), \( F \): ?, 1, 0, 1, ...
The output \( F \) toggles every two clock cycles after the initial state, so its frequency is 6 MHz.

Given the discrepancy with the options for Control = 0, let's reconsider if there's a specific initial condition or a non-standard behavior assumed. If the circuit somehow oscillates at 4 MHz when Control = 0, then option (A) would be plausible. However, based on standard analysis, Control = 0 leads to a stable output.
Assuming the intended answer aligns with one of the options despite the apparent contradiction: Option (A) has 6 MHz for Control = 1, which is consistent with our analysis. If Control = 0 somehow results in 4 MHz, then (A) would be the answer. Without further information or clarification on non-standard behavior, this remains speculative.
Final Answer: (A)