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To design a synchronous counter having sequence ( 0-1-0-2-0-3 ) and then repeats. To implement this counter is the minimum number of J-K flip-flops required. Note: This question was asked as a Numerical Answer Type.
  • a)
    1
  • b)
    2
  • c)
    4
  • d)
    5
Correct answer is option 'C'. Can you explain this answer?
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To design a synchronous counter having sequence ( 0-1-0-2-0-3 ) and th...
Designing a Synchronous Counter with Sequence (0-1-0-2-0-3)

Introduction:
A synchronous counter is a type of digital circuit that counts in a predetermined sequence and is synchronized with a clock signal. In this question, we are tasked with designing a synchronous counter that follows the sequence (0-1-0-2-0-3) and then repeats.

Understanding the Sequence:
To design the synchronous counter, we need to analyze the given sequence. Let's break down the sequence into binary representations:

- 0: 00
- 1: 01
- 2: 10
- 3: 11

From the binary representations, we can observe that we need at least 2 bits to represent the numbers 0 to 3.

Calculating the Number of Flip-Flops:
The number of flip-flops required in a synchronous counter is determined by the number of bits needed to represent the desired sequence. In this case, we require 2 bits.

Each flip-flop can store 1 bit of information. Therefore, we need 2 flip-flops to store the 2 bits required for the sequence.

Conclusion:
To implement the synchronous counter with the sequence (0-1-0-2-0-3), a minimum of 2 J-K flip-flops is required.

Note: It is important to note that the given sequence (0-1-0-2-0-3) does not repeat indefinitely. If the question stated that the sequence repeats indefinitely, we would need to consider additional flip-flops to accommodate the repeating pattern.
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Community Answer
To design a synchronous counter having sequence ( 0-1-0-2-0-3 ) and th...
Total 4. 2 J-K flip-flops for synchronous counter + 2 J-K flip-flops to make 2 bit counter. Actually, when we have repeated again, then after three, we don't know that our Synchronous counter will go to which zero. To make it work right, we need to move it to 1st zero after 3 2nd zero after 1 3rd zero after 2 i.e. 0 -> 1 -> 0 -> 2 -> 0 -> 3 (from here it again go to 1st zero). In order to decide which zero to move on, we use counters from 1 to 3; I have attached the truth table of a normal 2 bit synchronous counter using JK flip flop.
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To design a synchronous counter having sequence ( 0-1-0-2-0-3 ) and then repeats. To implement this counter is the minimum number of J-K flip-flops required. Note: This question was asked as a Numerical Answer Type.a)1b)2c)4d)5Correct answer is option 'C'. Can you explain this answer?
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