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

already seen their comments, referred books, NPTEL video lecture then saying it is an ambiguous question. There is indeed no restriction on the assignment of LSB & MSB.

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@Nitesh Singh 2 see the dates below their comments, references are given in 2018 and their comments pertaining to  "ambiguity" in question are of 2015....

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We Not need to solve this question by making entire diagram with JK flip flop it is very time consuming  just solve like below answwer last image 

https://gateoverflow.in/8219/gate-cse-2015-set-1-question-20?show=420352#a420352

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

Ref:

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these line of Mano specifies A) cannot be answer

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Agree

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@srestha
so here we should forget about that taking which one is MSB & which one is LSB concept, right??
we should only focus on the features of Johnson counter which is specified in Morris Mano.
Is this the approach of this question???

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@ MRINMOY_HALDER  i think we can because if  take E as MSB and A as LSB  and start from(0000) then the ans will be same following the features of Johnson counter

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srestha mam

If it was ring counter then it's like 0-0-0-0......

right???

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I havenot got u

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If ring counter initial state with 000 & no external i/p given then

000 - 000 -000 - 000.........

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@MRINMOY_HALDER

but why?? 

it can take 000-100-010-001-000

like that.

why only 0's?

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last FF output is fed to 1st FF i/p, that's why I'm thinking like this.

if initial state is 001 then it will be 001 - 100 - 010 - 001 - 100.....

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Advantage of Johnson counter(Twisted ring) over ring counter:

1) Twisted ring counter can count upto 2n states where as ring counter can count upto only n states.

2) Ring counter needs initialization. Twisted ring counter does not need any initialization (it will start counting in whichever state it is in.)

Ring counter with states 0000 without initialization will be stuck in 0000.

Reference: Youtube

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which one is the msb and which one is the lsb… it is not mentioned here, so how can we justtfy?

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A can also be the answer?

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@Pranavpurkar How?

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@Abhrajyoti00

see this → https://www.geeksforgeeks.org/differences-between-synchronous-and-asynchronous-counter/?ref=rp#:~:text=Synchronous%20Counter%20will%20operate%20in%20any%20desired%20count%20sequence.

count sequence does not matter in synchronous counter.

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@Pranavpurkar The line written is : “Synchronous Counter will operate in any desired count sequence.”

It means it doesn’t have to be always fixed (up/down). But it must operate in the desired count sequence and the desired count sequence is $0, 8, 12, 14, 15, 7, 3, 1, 0$ which will come if you make the truth table like this one: https://gateoverflow.in/8219/gate-cse-2015-set-1-question-20?show=184715#a184715

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@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 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 .