Consider the following circuit. What is the frequency at Q?
Consider the following circuit. What is the frequency at Q?
Answer: C. 0.5 MHz — Short answer: 0.5 MHz Explanation: Write the logic seen in the diagram. The gate receiving the AND output and the feedback is an exclusive-OR behaviour, so…
- A.
1 MHz
- B.
2 MHz
- C.
0.5 MHz
- D.
None of these
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Correct answer: C
Short answer: 0.5 MHz
Explanation:
Write the logic seen in the diagram. The gate receiving the AND output and the feedback is an exclusive-OR behaviour, so the next value of Q is Q ⊕ (A ∧ ¬B).
Define the pulse that drives the toggling: P = A ∧ ¬B. From the provided waveforms, A is a 1 MHz square wave and B is arranged so that the AND output produces pulses that occur in step with A but only during the intervals when B is low.
Each pulse of P causes the XOR feedback to flip (toggle) Q. When a signal toggles a bistable output on every active pulse, the output frequency is half the pulse rate (one full output cycle requires two toggle events).
Therefore, with A at 1 MHz and the gating shown, the pulses that cause toggling produce Q at half that rate: fQ = 1 MHz / 2 = 0.5 MHz.
Key takeaway: when feedback implements a toggle (XOR) controlled by a pulse train, the output toggles on each pulse and its frequency is half the pulse frequency.