GATE CSE 2015 | Set 1 | Question: 20

Question

Consider a $4$-bit Johnson counter with an initial value of $0000.$ The counting sequence of this counter is 

• A. $0, 1, 3, 7, 15, 14, 12, 8, 0$
• B. $0, 1, 3, 5, 7, 9, 11, 13, 15, 0$
• C. $0, 2, 4, 6, 8, 10, 12, 14, 0$
• D. $0, 8, 12, 14, 15, 7, 3, 1, 0$

Comments

@https_guru isn't twisted ring counter and jhonson counter same .
k bit cout 2k states .

 

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@manas_pant Not completely....The 2k states generated by the twisted ring counters are connected to 2k 2-input AND gates(Johnson Counters) to generate those sequence that doesn't violate the property Straight ring counter, i.e. At a time only one of the bit is 1. 

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1. Answer will Not be option A. Correct answer is Only option D.

2. Watch this lecture to understand Johnson Counter: Johnson Counter - Definition, Implementation

3. NOTE: The following statement is a Misconception & it is Wrong:

In case of synchronous counters, choosing MSB and LSB does not matter

In case of synchronous counters too, choosing MSB and LSB matters. If we change the MSB/LSB assignment, the counting sequence in decimal may change, even though the actual flip-flop states remain unchanged.

4. Difference between Johnson Counter & Twisted Ring Counter, Explained Here: https://youtu.be/vrWqcT9V7yQ

NOTE: Johnson Counter & Twisted Ring Counter are Not exactly same, But some authors use these terms interchangeably. Watch this to understand the difference.

Answer 1

$\text{Johnson Counter}$ is a switch‐tail ring counter in which a circular shift register with the complemented output of the last flip‐flop connected to the input of the first flip‐flop.

(D) is the correct answer!

Comments

@Abhrajyoti00

but this comment makes sense  https://gateoverflow.in/8219/gate-cse-2015-set-1-question-20?show=14588#c14588

as it also depends on selection of MSB and LSB.

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@Pranavpurkar https://gateoverflow.in/8219/gate-cse-2015-set-1-question-20?show=249304#c249304

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@Abhrajyoti00

Yes, i understood that part.

can you please point out whats wrong in this , https://gateoverflow.in/8219/gate-cse-2015-set-1-question-20?show=14588#c14588

Answer 2

option D

0000 - 0

1000 - 8

1100 - 12

and so on.

http://en.wikipedia.org/wiki/Ring_counter

Comments

If the answer would have been asked for ring counter then the answer is A. The johnson counter is switch‐tail ring counter so the complement of the output is given as input so answer is D. It would act like T-FF. Am i right?

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@Pooja Palod did you apply for correction in gate official key in 2015....

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@Shamim Ahmed NO

 If the answer would have been asked for ring counter then the answer would always be 0000.

Answer 3

Counting procedure of a Johnson's Counter

1. Starting from all 0's , each shift operation inserts 1's from the left until the register is filled with all 1's.
2. When the register has all 1s,each shift operation inserts 0's from the left until the register is filled with all 0's.
3. Go to step 1.

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0000 $\rightarrow$ ${\color{Red} 1}$000 $\rightarrow$ ${\color{Red} 1}{\color{Red} 1}$00 $\rightarrow$  ${\color{Red} 1}{\color{Red} 1}{\color{Red} 1}$0 $\rightarrow$  ${\color{Red} 1}{\color{Red} 1}{\color{Red} 1}{\color{Red} 1}$ $\rightarrow$  ${\color{Blue} 0}{\color{Red} 1}{\color{Red} 1}{\color{Red} 1}$ $\rightarrow$ ${\color{Blue} 0}{\color{Blue} 0}{\color{Red} 1}{\color{Red} 1}$ $\rightarrow$  ${\color{Blue} 0}{\color{Blue} 0}{\color{Blue} 0}{\color{Red} 1}$ $\rightarrow$ ${\color{Blue} 0}{\color{Blue} 0}{\color{Blue} 0}{\color{Blue} 0}$

OR

0 $\rightarrow$ 8 $\rightarrow$ 12 $\rightarrow$ 14 $\rightarrow$ 15 $\rightarrow$ 7 $\rightarrow$ 3 $\rightarrow$ 1 $\rightarrow$ 0

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$\therefore$ Option D is correct answer.

Comments

Although this is the most appropriate answer to this question still I have one query.

Consider D flipflops in johnson counter are in series Q0Q1Q2Q3 from left to right and now we can either assign LSB of the binary no. to Q0 or LSB of the binary no. to Q3. But how you would be so sure to assign LSB to Q3 only? We could assign it to any one of them for sure. There is no restriction.

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you insert data in anyway but working of Johnson's counter will not change right ?

It will be same as i have written in the answer.

So suppose instead of entering 1000 you entered 0001,

then it will start generating like  0001->0000 -> 1000.... and so on. i.e. 1->0->8

But i know that the output should be 1000-> 1100 -> 1110 .... i.e. 8 ->12 -> 14

So by looking at the pattern i would be sure that i have written entered data incorrectly.

NOTE :-

We are just saying that leftmost is MSB and rightmost is LSB but who cares. You can treat the input and outputs whichever way you want. I can say that i will treat my MSB as righmost bit and leftmost bit as LSB and accordingly assume my inputs and outputs produced in decimal format. i.e. 0001 means 8 to me.

Answer 4

Johnson counter /twisted ring counter :-

"N" bit then there will be "2N" states 

Option B easily eliminated as it has 9 state form rest we will have to check out.

Answer :- D 

 

Answer 5

Johnson counter is ring counter hence after period of clock cycles its output will repeat .

here 0000 is given as initial state and hence 1 is given as preset in first flipflop and then this one is propagated to consecutive flipflops producing

1000

1100

1110

1111

0111

0011

0001

0000

as output .

Answer 6

 Straight Ring Counter. 

there are 4 states in a 4-bit Ring Counter.

 

Johnson Counter (Twisted Ring Counter)

So answer is option D

Reference : Ring Counter in Digital Logic - GeeksforGeeks