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37 votes
37 votes

The following arrangement of master-slave flip flops

has the initial state of $P, Q$ as $0, 1$ (respectively). After three clock cycles the output state $P, Q$ is (respectively),

  1. $1, 0$
  2. $1, 1$
  3. $0, 0$
  4. $0, 1$
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5 Answers

Best answer
60 votes
60 votes

Here, clocks are applied to both flip flops simultaneously. Outputs for $3$ cycles will proceed as follows:

  • When $11$ is applied to $JK$ flip flop it toggles the value of $P.$ So, output at  $P$ will be $1.$
  • Input to D flip flop will be $0$ (initial value of P).  So, output at $Q$ will be $0.$

    $JK$ flip flop it toggles the value of $P.$ So, output at  $P$ will be $0.$

  • Input to D flip flop will be $1$ (current value of P) . So, output at $Q$ will be $1.$
  • $JK$ flip flop it toggles the value of $P.$ So, output at  $P$ will be $1.$
  • Input to D flip flop will be $0$ (initial value of P)  so output at $Q$ will be $0.$

So, answer is A.

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20 votes
20 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].

4 votes
4 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
3 votes
3 votes

-> As there is Master Slave configuration, 1st flipflop will respond to the positive edges and the 2nd flipflop to the negative edges.

-> As the given Flipflops are synchronous,(clock is simultaneously given to both FFs) $Q$ responds according to immediate values of $P$.

-> Thus, after 3 cycles, as shown in the diagram, $P = 1$ and $Q=0$ 

Answer:

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