Consider a 4-bit Johnson counter with an initial value of 0000. The counting sequence of this counter is
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GATE CSE 2015 Set 1 · Q20
Time
Select one answerSingle correct
Solution
Official answer: D) 0, 8, 12, 14, 15, 7, 3, 1, 0
Combinational and Sequential CircuitsA 4-bit Johnson counter (or twisted-ring counter) with right-shift operation shifts bits to the right and feeds the complement of the last bit back into the first position:
Initial state:
Invert LSB () and shift right:
Invert LSB () and shift right:
Invert LSB () and shift right:
Invert LSB () and shift right:
Invert LSB () and shift right:
Invert LSB () and shift right:
Invert LSB () and shift right:
Invert LSB () and shift right:
Thus, the decimal counting sequence is 0, 8, 12, 14, 15, 7, 3, 1, 0, corresponding to option (D).
Try the question first, then check your work.
Examiner trap
In a right-shifting Johnson counter, the inverted output of the last flip-flop is fed back into the input of the first flip-flop.
This Combinational and Sequential Circuits previous year question is Q20 of the GATE CSE 2015 Set 1 paper. Find more GATE CSE PYQs on Combinational and Sequential Circuits in Digital Logic, or browse all GATE CSE PYQs.
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Combinational and Sequential Circuits has 32 GATE CSE PYQs across 22 papers from 2011 to 2026. 8 of them are 1-mark MCQs like this one, and GATE CSE 2015 Set 1 has 2.
How GATE tests Combinational and Sequential CircuitsIndependent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.