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GATE | GATE-CS-2015 (Set 1) | Question 47

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A positive edge-triggered D flip-flop is connected to a positive edge-triggered JK flipflop as follows. The Q output of the D flip-flop is connected to both the J and K inputs of the JK flip-flop, while the Q output of the JK flip-flop is connected to the input of the D flip-flop. Initially, the output of the D flip-flop is set to logic one and the output of the JK flip-flop is cleared. Which one of the following is the bit sequence (including the initial state) generated at the Q output of the JK flip-flop when the flip-flops are connected to a free-running common clock? Assume that J = K = 1 is the toggle mode and J = K = 0 is the state-holding mode of the JK flip-flop. Both the flip-flops have non-zero propagation delays.

(A)

0110110…

(B)

0100100…

(C)

011101110…

(D)

011001100…



Answer: (A)

Explanation:

 

Initially Q output of D – FF  = 1 Initially Q output of JK – FF  = 0 Now with the help of present state and next state table we can see what is happening in circuit.

  • Toggle: J = K = 1
  • Hold : J = K = 0

 

We see from table Q output of D-FF is going to next state input of JK-FF and the bits sequence produced is like 110110….. Including initial condition (0) we get output as 0110110110…. Hence answer is (A) part. Another Explanation: Here, it is given that JK flip flop will toggle when J = K = 1 and it will retain the output if J = K = 0. Also, the output of the D flip flop would remain the same as the input. So, we have Initial output : D = 1

JK = 0

After clock 1 : D = 0 (D gets 0 as input 
from initial output of JK, so output = 0)

JK = 1 (J = K = 1 from initial output of D, so output would be toggled from 0 to 1)

After clock 2 : D = 1 (D gets 1 as input 
from previous output of JK, so output = 1)

JK = 1 (J = K = 0 from previous output of D, so output would be retained to 1)

After clock 3 : D = 1 (D gets 1 as input
from previous output of JK, so output = 1)

JK = 0 (J = K = 1 from previous output of D, so output would be toggled from 1 to 0)

After clock 4 : D = 0 (D gets 0 as input
from previous output of JK, so output = 0)

JK = 1 (J = K = 1 from previous output of D, so output would be toggled from 0 to 1)

After clock 5 : D = 1 (D gets 1 as input
from previous output of JK, so output = 1)

JK = 1 (J = K = 0 from previous output of D, so output would be retained to 1)

Thus, the bit sequence generated at the Q output of the JK flip flop will be 0110110…  


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Last Updated : 29 Sep, 2021
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