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+17 votes

The following arrangement of master-slave flip flops


has the initial state of P, Q as 0, 1 (respectively). After a clock cycle the output state P, Q is (respectively),

  1. 1, 0
  2. 1, 1
  3. 0, 0
  4. 0, 1


asked in Digital Logic by Veteran (68.8k points)
edited by | 1.8k views
after how many clock cycles?

3 Answers

+24 votes
Best answer
Here clocks are applied to both flip flops simultaneously

When 11 is applied to jk flip flop it toggles the value of P so op at  P will be 1

Input to D flip flop will be 0( initial value of P)  so op at Q will be 0

So ans is a
answered by Veteran (33.8k points)
edited by
JK-11 and previous state of P is 0, so next state of P will be 1. right?
but he is given to assume as master slave did it mean i need negation of the clock to the D-flip flop??

plz explain i am confused

@Arjun Sir,
IF the question were to calculate the output after 3 clock cycles  . Would the answer be 1,1...?
IS my approach correct ?

PC how do you get p=1 in cyle 2 when j=1,k=1 ,toggle happens it means previously p=1 after toggle p=0 na

u may get 01,10,01,10 ............
Yeah , u r correct . I did a mistake .

thx :)
It is given as a master slave FF then why the clock is not inverted for second FF?
+9 votes

now when clock is aplied J= 1 and K=1 then (it do complement of P) i.,e output at p = 1 and at same time FF D has input 0 so it change Output Q as ) only [ on D-FF input = output]

SO output will be p= 1 and Q= 0 

[ keep in mind both FF work at same time dont wait for 1st complete then second will start].

answered by Veteran (57.5k points)

@sittian  hopw you got it now.

You are not getting this ia synchronous ckt I.e. clock is not present on any FF . Both FF work at together here .. Check clock input properly..

Input to D comes from JK FF not
+2 votes
first find the next state for jk-ff and d-ff  which is p(n+1)=not(p) and q(n+1)=d=p
given in question p=0 and q=1 so next state for p=1 and q=0
so a is ans
answered by Boss (5k points)

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