The three-stage Johnson counter as shown in figure is clocked at a constant frequency of ƒc from the starting state of Q2Q1Q0 = 101. The frequency of output Q2Q1Q0 will be
ƒc/8
ƒc/6
ƒc/3
ƒc/2
Correct Option: D
We see that 1 0 1 repeat after every two cycles, hence frequency will be ƒc/2 .