ISRO 2015 - Johnson counter [EE]

Question

A three stage Johnson counter ring in figure is clocked at a constant frequency of fc from starting state of Q0Q1Q2 = 101. The frequency of output Q0Q1Q2 will be

(a) fc / 2
(b) fc /6

(c) fc/ 3

(d) fc /8

Comments

please someone verify the answer.

Answer 1

It only Count two stages  101 and 010 

hence f_c/2

Comments

Irrespective of states in will goes for no of flipflops. so, i think f/2n = f/6

Answer 2

Johnson counter is mod 2N counter where N is number of FF's. The output frequency will be fc/2N = fc/6 (option B)

Comments

I think your answer is valid when the initial value is 000. Here, the initial value is 101. And thus the counter is a mod-2 counter 101->010. So, the answer should be fc/2. Someone, please verify.

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@Vishal Goel sir,  official key answer given is f/6.