GATE 2015 EE (EE2) – Question 47
Analog and Digital Electronics · VCOs and timers, combinatorial and sequential logic circuits, multiplexers, demultiplexers · 2 marks · Multiple choice
The sequential circuit starts at Q1Q0=00. Its state immediately after the 333rd clock pulse is
- 00
- 01
- 10
- 11
Practise this question in The GATE Grind →
Show answer and explanation
Correct answer: (B) 01
Explanation
From the wiring, the first JK pair has J0=Q̄1,K0=Q1, making Q0,next=Q̄1. The second has J1=Q0,K1=Q̄0, making Q1,next=Q0.
The state cycle is therefore 00→01→11→10→00. Its period is four. Since 333 mod 4=1, the state is the first after 00: **01, B**.
More VCOs and timers, combinatorial and sequential logic circuits, multiplexers, demultiplexers questions
- GATE 2015 EE (EE2) Q48 – Implement f(A,B,C,D)=(1,5,12,15) using an 8:1 MUX. A is MSB and ABC drive S2,S1,S0. Which…
- GATE 2026 EE Q45 – The digital circuit shown has 3 inputs (x, y, and z). The simplified logical expression…
- GATE 2026 EE Q46 – The MOSFET switches shown in the circuit are ideal. Which of the following is the correct…
- GATE 2025 EE Q37 – A Boolean function is given as f=( u+ v+ w+ x)( u+ v+ w+x)( u+v+ w+ x)( u+v+ w+x) The…
- GATE 2024 EE Q24 – Simplified form of the Boolean function F(P,Q,R,S)= P Q+ PQS+P Q R S+P QR S is
- GATE 2024 EE Q25 – In the circuit, the present value of Z is 1. Neglecting the delay in the combinatorial…
All VCOs and timers, combinatorial and sequential logic circuits, multiplexers, demultiplexers questions →