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),
Yes, you are right that in asynchronous counter the output of 1FF is given to other FF but the output is given as the Clock of other FF while in the synchronous we have same clock.
Please Correct me if I am wrong.
Here, clocks are applied to both flip flops simultaneously. Outputs for $3$ cycles will proceed as follows:
$JK$ flip flop it toggles the value of $P.$ So, output at $P$ will be $0.$
So, answer is A.
@Arjun Sir, IF the question were to calculate the output after 3 clock cycles . Would the answer be 1,1...? IS my approach correct ?
P$_{N}$ = $\sim$P
Q$_{N}$ = P
$P_{Next} = JP'+K'P = 1P'+1'P = P'$ $Q_{Next} = D = P$
State Diagram:
So, the correct answer is $(A)$.
@Doley
In general $:Q_{Next} = J\:\overline{Q} + \overline{K}\:Q$
In the first clock itself when the clock is high:
As we know that clock time is generally smaller than the delay due to FF, when the output of J-K becomes 1 before the clock goes low, the output of D will change to 1. So at the end of cycle one, the output seems to become 1 1.
Similarly after clock cycle two and three output will become: 0 0 and 1 1 respectively.
@Lakshman Bhaiya kindly correct me.
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].
@sittian hopw you got it now.
-> 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 $Q=0$
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