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+1 vote
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Consider the logical circuit shown below:

If initially ABC = 000 then after how many clock pulses the circuit will reach its initial stage?

a) 5

b) 6

c) 7

d) 8

in Digital Logic by Loyal (7.2k points)
edited by | 221 views
0

Here is what i am getting the pattern,

CLK   ABC
0 000
1 100
2 110
3 111
4 011
5 001
6 000

What will be answer 5 or 6 ?

0
6 is correct

1 Answer

+4 votes
Best answer

$J_A= C' , K_A = C ,$ So, $A^+ = J_AA'+K_A'A = C'A'+C'A = C'$
$D_B = A,$ So $B^+ = D_B = A$

$J_C =B , K_C = B' ,$ So $C^+ = J_CC'+K_C'C = BC'+BC = B$

So,$A^+=C', B^+ = A,$ and $C^+ = B$

Initially $ABC= 000$  so counter goes as 

$000$

$100$

$110$

$111$

$011$

$001$

$000$

After $6$ clocks it will be back to initial state

Note:

1. Characterstic equation of JK FF , $Q_{t+1}=JQ_t'+K'Q_t$ and  of D FF, $Q_{t+1}=D$

2. It is a Johnson counter , having $n$ flip-flops or ($n$-bit Johnson counter), and  having $2n$ states.

3. $n$-bit Johnson counter is also known as mod$2n$ counter.

by Veteran (56.8k points)
selected by
0
Thank you very much sir.

I have one more doubt, Johnson Counter can be made by only all JK - FF or all D - FF but here they have used both. Is this one also be called as Johnson counter ?
+1

Any thing designed by $x$ FF can be designed by $y$ FF also
One FF can be converted into another equivalent FF .
There is a property of Johnson counter, that is switch tail , complement of last become next state of First FF, and rest are just shifted , as here $A^+ = C', B^+= A, C^+= B $

0
thanks :)
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