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29 votes

Consider a combination of $\text{T}$ and $\text{D}$ flip-flops connected as shown below. The output of the $\text{D}$ flip-flop is connected to the input of the $\text{T}$ flip-flop and the output of the $\text{T}$ flip-flop is connected to the input of the $\text{D}$ flip-flop.

Initially, both $Q_{0}$ and $Q_{1}$ are set to $1$ (before the $1^{\text{st}}$ clock cycle). The outputs

  1. $Q_{1}Q_{0}$ after the $3^{\text{rd}}$ cycle are $11$ and after the $4^{\text{th}}$ cycle are $00$ respectively.
  2. $Q_{1}Q_{0}$ after the $3^{\text{rd}}$ cycle are $11$ and after the $4^{\text{th}}$ cycle are $01$ respectively.
  3. $Q_{1}Q_{0}$ after the $3^{\text{rd}}$ cycle are $00$ and after the $4^{\text{th}}$ cycle are $11$ respectively.
  4. $Q_{1}Q_{0}$ after the $3^{\text{rd}}$ cycle are $01$ and after the $4^{\text{th}}$ cycle are $01$ respectively.
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7 Answers

Best answer
35 votes
35 votes

Since it is synchronous so, after every clock cycle T will toggle if input is 1 and will be in Hold State if input is 0. D flip-flop’s output always follows input

  • After $1$ clock cycles $:Q_{1} = 0$ (Toggle)  $Q_{0} = 1$
  • After $2$ clock cycles $:Q_{1} = 1$ (Toggle) $Q_{0} = 0$
  • After $3$ clock cycles $:Q_{1} = 1$ (Hold)  $Q_{0} = 1$
  • After $4$ clock cycles $:Q_{1} = 0$ (Toggle) $Q_{0} = 1$

Hence, option (B) is correct.

edited by
43 votes
43 votes

Answer: (B)

2 votes
2 votes
We can also solve it in less time using the characteristic equation of D and T flip flop.

We denote T FF output by Q1 and D FF output by Q0.

Now , Q1N = Q1 EXOR T = Q1 EXOR Q0

 and Q0N= D = Q1

Initailly Q1=Q0=1

Q1     Q0        Q1N     Q0N

1       1            0           1          1st clock cycle

0        1         1            0              2nd

 1       0            1           1                    3rd

1          1        0            1                        4th

Hence the answer -     11,O1    option B
2 votes
2 votes

Option (B) is correct

Answer:

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