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+33 votes

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

- 0, 1, 3, 7, 15, 14, 12, 8, 0
- 0, 1, 3, 5, 7, 9, 11, 13, 15, 0
- 0, 2, 4, 6, 8, 10, 12, 14, 0
- 0, 8, 12, 14, 15, 7, 3, 1, 0

+20

In case of synchronous counters choosing msb and lsb does not matter , it can be from any side because the operation is done in parallel per clock pulse but in case of asynchronous counter it matters because it is done serially . Since johnson counter is synchronous it wont matter here so answer can be both a and d.

+1

Both options $A$ and $D$ are actually reverse orders of each other. So, if one can be done other can also be done by changing the order of MSB and LSB.

0

No matter how much you try to justify Answer- D but this is indeed not a proper question asked.

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.

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.

0

If you think the question is ambiguous** **because of ambiguity between MSB and LSB see the reference provided by Arjun sir and Srestha mam.

0

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.

0

@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....

+17 votes

Best answer

0

@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???

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???

0

@ 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

0

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.....

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

0

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

+19 votes

+5

why not (A)....?? if i consider q3 as MSB ... i got option (a) ... and here not mention what is MSB and LSB ... or

i consider D0 = Q3' , D1=Q0, D2 =Q1, D3= Q2

Q3N =D3, Q2N= D2 , Q1N= D1, Q0N= D0.

i consider D0 = Q3' , D1=Q0, D2 =Q1, D3= Q2

Q3N =D3, Q2N= D2 , Q1N= D1, Q0N= D0.

+3

Arjun Sir I marked A in exam...lost marksin wiki link it is given that clocl should be applied to Msb...but i dont think there is any such rule

0

The answer can be either A or D depending on which one you choose as MSB since nothing is given .

The persons answering it either A or D should be given mark.

The persons answering it either A or D should be given mark.

+3 votes

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 .

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 .

+3 votes

**Counting procedure of a Johnson's Counter**

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

--------------------------------------------------------------------------------------------------------------------------------------------------------------------

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

--------------------------------------------------------------------------------------------------------------------------------------------------------------------

$\therefore$ Option D is correct answer.

+1

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.

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.

0

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.

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