The three stage Johnson counter as shown in figure below is clocked at a constant frequency of $15 M H z$ from the starting state of $Q_{2} Q_{1} Q_{0} = 101$ . The frequency of output $Q_{2} Q_{1} Q_{0}$ will be _____ $M H z .$

7.5

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Solution

The correct option is A 7.5

$J_{2}$ $K_{2}$ $J_{1}$ $K_{1}$ $J_{0}$ $K_{0}$ $Q_{2}^{+}$ $Q_{1}^{+}$ $Q_{0}^{+}$

1 0 1

0 1 1 0 1 1 0 1 0

1 0 0 1 0 0 1 0 1

0 1 1 0 1 1 0 1 0

1 0 0 1 0 0 1 0 1

101 repeated after every two cycles,
hence frequency of the output will be $\Rightarrow \frac{15}{2} = 7.5 M H z .$

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$J_{2}$	$K_{2}$	$J_{1}$	$K_{1}$	$J_{0}$	$K_{0}$	$Q_{2}^{+}$	$Q_{1}^{+}$	$Q_{0}^{+}$
						1	0	1
0	1	1	0	1	1	0	1	0
1	0	0	1	0	0	1	0	1
0	1	1	0	1	1	0	1	0
1	0	0	1	0	0	1	0	1

The three stage Johnson counter as shown in figure below is clocked at a constant frequency of 15MHz from the starting state of Q2Q1Q0=101. The frequency of output Q2Q1Q0 will be _____MHz. 7.5

The correct option is A 7.5 J2 K2 J1 K1 J0 K0 Q+2 Q+1 Q+0 1 0 1 0 1 1 0 1 1 0 1 0 1 0 0 1 0 0 1 0 1 0 1 1 0 1 1 0 1 0 1 0 0 1 0 0 1 0 1 101 repeated after every two cycles, hence frequency of the output will be ⇒152=7.5MHz.

The three stage Johnson counter as shown in figure below is clocked at a constant frequency of $15 M H z$ from the starting state of $Q_{2} Q_{1} Q_{0} = 101$ . The frequency of output $Q_{2} Q_{1} Q_{0}$ will be _____ $M H z .$

7.5