3.8k views

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

1. 0, 1, 3, 7, 15, 14, 12, 8, 0
2. 0, 1, 3, 5, 7, 9, 11, 13, 15, 0
3. 0, 2, 4, 6, 8, 10, 12, 14, 0
4. 0, 8, 12, 14, 15, 7, 3, 1, 0
edited | 3.8k views
+14
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.
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@mcjoshi for asynchronous counter what will be answer???

$\text{Johnson Counter}$ 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.

selected
+2

Ref:

+1

these line of Mano specifies A) cannot be answer

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Agree

option D

0000 - 0

1000 - 8

1100 - 12

and so on.

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

+3
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.
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option A and D both are correct .
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Isn't MSB defined in Johnson counter?
+1
as per i know you can take any one as msb ( Q0 or Q3 ) , its depends on the user .
+1
I have to see that, but GATE key had only D.
+1

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

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yes sir in answer key ans is (D)... but we can do any method ....
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???
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@csegate2 nice , its a gif right ?? is it made by you ? or same old policy ctrl c + Ctrl v  :P
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i dont know who you are , but I will find you, I will learn the secret of that pic :D ....
+1

Listen everyone, I'll kill you I'm saying

for mithilesh

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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.
+5

What is the mistake in this? please explain.

I think correct answer is A

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u r missing 14 only . rest is OK.
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The flip flops should be arranged like D3->D2->D1->D0, in the same sequence as that of the table.
0

from morris mano

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 .

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.

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