A three stage Johnson counter ring in figure is clocked at a constant frequency of fc from starting state of Q0 Q1 Q2 = 101. The frequency of output Q0 Q1 Q2 will be [Figure description: Three JK flip-flops in a row (outputs Q0, Q1, Q2 with complements), all driven by a common CLK line. Each flip-flop's J input is connected to the Q output of the previous stage's... specifically J1/K1 inputs are connected from the previous stage's Q and Q-bar outputs (J from Q, K from Q-bar) for stages 2 and 3, and the last stage's outputs Q2-bar and Q2 are fed back along a long wire over the top to the first flip-flop's J and K inputs (J0 from Q2-bar, K0 from Q2), forming a twisted-ring (Johnson) counter.]
- A. fc / 2
- B. fc / 6
- C. fc / 3
- D. fc / 8
Answer: B. fc / 6
Explanation: A 3-stage Johnson counter has 2n = 6 states, so output frequency is fc/6.