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GATE1993-6-3

31 votes
6k views

Multiple choices can be correct. Mark all of them.

For the initial state of $000$, the function performed by the arrangement of the $J-K$ flip-flops in figure is:

 

  1. Shift Register
  2. $Mod- 3$ Counter
  3. $Mod- 6$ Counter
  4. $Mod- 2$ Counter
  5. None of the above
in Digital Logic
edited by 6k views

9 Answers

48 votes
 
Best answer

Circuit behaves as shift register and mod6 counter

$$\begin{array}{c|c}\hline \textbf{Clock Cycle}  &  \textbf{Output} \\\hline \text{1} & \text{100}\\ \text{2} & \text{110}\\ \text{3} & \text{111}\\ \text{4} & \text{011}\\ \text{5} & \text{001}\\ \text{6} & \text{000}\\\hline \end{array}$$

This is Johnson counter which is an application of Shift Register. And Johnson counter is mod 2N counter.


edited by
0

can you please explain how you got the states(100,110,111,011,001,000)

initial state is 000 so for the clock 1

                i am getting                     Flip flop 1:  J =1(Q3`)  k=0(Q3)   Q1=1

                                                     Flip flop 2: J=1(Q1) k=0(Q1`)    Q2=1

                                                     Flip flop 3: J=1(Q2) k=0(Q2`)     Q3=1

clock 2

                                     Flip flop 1:  J =0(Q3`)  k=1(Q3)          Q1=0

                                    Flip flop 2: J=0(Q) k=1(Q`)                Q2=0

                                    Flip flop 3: J=0(Q)  k=1(Q`)                Q3=0

                                     

same repeating .. 

11
Murali see what u r doing is not synchronous. the clock is provided to all jk flipflops at the same time so ->

assuming the arrangement is Q1  Q2  Q3

Q1     Q2   Q3

0       0      0     <-init

1       0      0     (q1 is 1 as -q3 gets attached to j1, q2 is zero becoz at the same time q1(which is zero) gets attached to j2 and similarly q2(which is zero) gets attached to j3.

and similarly for rest....hope this helps
0
The given circuit is only Mod-6 counter, not Shift Register.. Can u please explain why u think it is a shift register
0
why shift register??
1
It is shift register, mod 6 counter. Ok. Why not mod 2 counter ? It does mod 2 counting too.
Should not the answer be A,C,D ?
0
yes, but u have to answer tha optimal solution. It is both mod 2 and mod 3. So, optimally we can say it is mod 6 counter
0
how it is mod3 ?? :O
0
Bcoz it is johson counter which is nothing but shift register
0
either its mod6 or mod 2 ..how mod 3 ?
0
Can you please draw the table for asynchronous clock also for the same initial values. So, that it will be easier to differentiate between the two
13
@Ahwan, In the question, the function of circuit is asked for initial state $000$. So we can never reach the remaining $2$ states that are $010$ and $101$ which are responsible for $mod \ 2$ counting.
If it the initial state were $010$ or $101$ then the answer would be Shift register and $mod \ 2$ counter.

edited by
0
Thq soo much dinesh
0
Johnson register only attached with Q' not Q so why you say it is Johnson register?
0
It is Johnson only. Isn't it?
18 votes
1st approach : Start from 000 and use conventional method (i.e. Truth table) to find modulus.

2nd approach : see, given circuit is nothing but Jonson Counter . Number of States in Jonson counter will be 2n where n is number of flip flops. here n is 3 so number of states are 6 . mod 6 counter..

edited by
0

Yes, It's twisted ring counter i.e Johnson Counter which does mod 2N counting.

Here, n=3 so, mod 6 counter. 

Johnson counters are application of Shift Registers.

17 votes

According to me, I think that A, C and D are the answer. Pleas, find the attached screen shot.

0

this is not D-FF. in JK FF, Q(next) = JQ' + K'Q then how can you consider that Q0N=Q1, Q1N=Q2....??

5

Raju Kalagoni When i/p's of JK-ff are complement to each other then it is D-ff.

0
@Raju Kalagoni, here I didn't show the part of find JK, (its my bad) since, in this case it will be same. But, better approach is to find the JK part for each J0K0, J1K1, and J2K2, otherwise we may not get the right answer. To be safer side we should find the JK part as well.
0
0
Initial state given is 000 so option D) is not possible!
7 votes

A. Shift Register 
C. Mod-6 Counter

Given circuit diagram is for a Mod-6 Johnson Counter.


edited by
0
@amar , it is johnson counter right ? Mod-6 counter.
0
No, that's different.
0
check this for your ref, same question

https://gateoverflow.in/17237/gate1993_6-3
1

this(link) is how a Mod 6 counter looks like. Anything that has 6 states does not imply that it's a Mod 6 counter.

Yes, it is a Johnson counter, but not Mod 6 counter. Also, it shifts the binary number $000111$ so, we are calling it a Shift register.


edited by
7

@amar there are lot of e - resources over the web, that support that we can say it is mod6 counter

" The sequence for a Mod N counter may follow the binary count or may be any other arbitrary sequence. "

Between, this is from M. Mano, Section 6.5, 5th edition, pg 278

1
Sir, I've updated it now.
1
In the table, you shows sequence of states, along with sequence complemented outputs (states) , with 3 FF we can have any (shift) register of 3 bits only. (not as shifting 000111)
0
Sir i think answer will be both shift register as well as the mod 6 counter. Right?
0
why not mod 2 counter?
0 votes
There are two counters present for the given circuit.

1. Mod 6 counter (States 000,001,011,111,110,100)

2. Mod 2 counter (As the remaining states 010 and 101 are also doing mod 2 counting.)

And the question is also saying multiple choices can be correct.

So answer should be C and D.
0 votes

option : C

please mention if any mistake is found..

0 votes
It is johnson counter of three bit. For Johnson counter I there is n bit then it will have 2n state =6 (here) so this is shift reg with mod 6
0 votes

It is "Johnson counter" which is MOD2n counter where "n" number of FF so for this example its "6" .....if you want to visualize how 6 coming then  draw a the table and see the  flow of state will be like this here its "shifting" from one state to another 

Answer:- Mod6, shift register


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