Difficulty: Easy
Correct Answer: D flip-flop
Explanation:
Introduction / Context:J–K flip-flops are versatile because appropriate input conditioning can emulate other flip-flop types. One common technique wires the inputs so the device samples a single data line D effectively, reproducing D flip-flop behavior. Recognizing this mapping simplifies design when only J–K elements are available.
Given Data / Assumptions:
Concept / Approach:From the J–K characteristic table: J = 1, K = 0 sets Q to 1; J = 0, K = 1 resets Q to 0; J = K = 0 holds; J = K = 1 toggles. With J tied to D and K tied to D̄, the pair (J,K) always forms either (1,0) when D = 1 or (0,1) when D = 0. Thus, at each valid clock edge, Q_next equals D, which is exactly the behavior of a D flip-flop.
Step-by-Step Solution:
If D = 1 → J = 1 and K = 0 → Q_next = 1 (set).If D = 0 → J = 0 and K = 1 → Q_next = 0 (reset).There are no cases with J = K simultaneously under this wiring.Therefore, the device copies the D input at the clock edge, i.e., it behaves as a D flip-flop.Verification / Alternative check:Implement the wiring in a simulator and apply a data waveform to D with clean edges relative to the clock. Observe Q tracks D on active edges exactly like a DFF. This is a classic textbook equivalence used in lab exercises.
Why Other Options Are Wrong:
Common Pitfalls:Failing to maintain D stable around the clock edge (setup/hold violations); miswiring complements leading to unintended toggling; leaving inputs floating, which can cause unpredictable behavior.
Final Answer:D flip-flop
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