Consider the partial implementation of a 2-bit counter using T flip-flops…
2004
Consider the partial implementation of a 2-bit counter using T flip-flops following the sequence 0-2-3-1-0, as shown below
To complete the circuit, the input X should be
- A.
Q2'
- B.
Q2 + Q1
- C.
(Q1 ⊕ Q2)'
- D.
Q1 ⊕ Q2
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Correct answer: D
Key insight: For a T flip-flop, the output toggles when T = 1 and remains the same when T = 0. Thus T = current_Q ⊕ next_Q.
Apply this to the least significant bit (LSB). List the current state (Q2 Q1), the next state, and the needed T for Q1:
State 00 → next 10: Q1 is 0 → 0, so T1 = 0
State 10 → next 11: Q1 is 0 → 1, so T1 = 1
State 11 → next 01: Q1 is 1 → 1, so T1 = 0
State 01 → next 00: Q1 is 1 → 0, so T1 = 1
So the required T1 values for states 00, 10, 11, 01 are 0, 1, 0, 1 respectively.
Compare this to the Boolean expression Q1 ⊕ Q2 evaluated for the same states:
00: Q1 ⊕ Q2 = 0
10: Q1 ⊕ Q2 = 1
11: Q1 ⊕ Q2 = 0
01: Q1 ⊕ Q2 = 1
Conclusion: The required input X for the LSB T flip-flop is Q1 ⊕ Q2.
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