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+39 votes

In the sequential circuit shown below, if the initial value of the output $Q_1Q_0$ is $00$. What are the next four values of $Q_1Q_0$?

- $11$, $10$, $01$, $00$
- $10$, $11$, $01$, $00$
- $10$, $00$, $01$, $11$
- $11$, $10$, $00$, $01$

+40 votes

Best answer

**Option A.**

$2^{nd}$ flip-flop will be active only when $1^{st}$ flip flop produces output $1.$ For clocks $2$ and $4$ old output is retained by Flip-Flop $2$.

$\begin{array}{c|c}T&Q_{n+1}\\\hline 0&Q_n\\1&\overline{Q_n}\end{array}$$\underset{\text{For rows 2 and 4, Clk for $T_1$ is 0 and hence old o/p is retained}}{\begin{array}{cc|cc}T_0&Q_{0}&T_1&Q_1\\\hline &0&&0\\\hline 1&1&1&1\\1&0&1&1\\1&1&1&0\\1&0&1&0\end{array}}$ $\begin{array}{c|c}Q_1&Q_{0}\\\hline1&1\\1&0\\0&1\\0&0\end{array}$

0

@neha_singh 2nd Flip-Flop will be active only when 1st Flip Flop produce output 1 . For clock 2 and 4 Old output is retained

0

see, clock of T1 will be triggered when Q_{0 }changes from 0-->1 as it is a positive edge triggered flip flop.

+23

Such kind of question make more sense if we solve using timing diagram.

Please note that the FF indicated in the diagram are positive edge triggered and are asynchronous.

0

A very subtle hint given here is that the circuit is a flip-flop, and not latch. I was assuming it to be a latch.

+2

Why are we considering present state of $Q_0$? shouldn't we take previous state of $Q_0$, for determining present state of $Q_1$ ??

+1

@Rishabh Gupta 2 that is the same doubt which is annoying me! Have you found the answer to your question?

+17 votes

answer - A

Initially Q0Q1 = 00

after first clock signal Q0 will be toggled and change to 1. It will activate clock of second FF in sequence and toggle that one as well.

Hence next output Q0Q1=11

Similarly next few sequences for Q0Q1 will be 01, 00

Hence answer sequence Q1Q0 will be 11,10,01,00

Initially Q0Q1 = 00

after first clock signal Q0 will be toggled and change to 1. It will activate clock of second FF in sequence and toggle that one as well.

Hence next output Q0Q1=11

Similarly next few sequences for Q0Q1 will be 01, 00

Hence answer sequence Q1Q0 will be 11,10,01,00

0

Can someone explain in detail ?

" after first clock signal Q0 will be toggled and change to 1 " - How is this possible ?

Logic 1 is constantly applying.. So how are you getting 01, 00 ?

Can you please explain it with boolean algebra ?

I am not getting the working

" after first clock signal Q0 will be toggled and change to 1 " - How is this possible ?

Logic 1 is constantly applying.. So how are you getting 01, 00 ?

Can you please explain it with boolean algebra ?

I am not getting the working

+3

0

Q0 Q1

0 0

1 1

0 1

Now , In first FF

1 XOR (last Q0 = 0 ) = 1

This activate 2 nd FF

1 XOR ( last Q1 = 1 ) = 0

So the states are 00 , 11 , 01 , 10

Am I correct ?

0 0

1 1

0 1

Now , In first FF

1 XOR (last Q0 = 0 ) = 1

This activate 2 nd FF

1 XOR ( last Q1 = 1 ) = 0

So the states are 00 , 11 , 01 , 10

Am I correct ?

+1

Activating the 2nd FF clock is ok but how can you get the o/p of 2nd FF in the same cycle? It should be the next +ve edge, when it should be updated (like we do in Up or Down counter, where the next bit's clock is dependant on the previous bit's edge..) So , as per that scene why cant be the o/p as 01,00,11... ? Please explain somebody....

0

I am also having same confusion. how can you get the o/p of 2nd FF in the same cycle?

Please explain somebody.

0

@pc this is +ve edge triggered flip flop and this flip flop will be activated when clock pulse will be 1 .???

is it rt na ?????

is it rt na ?????

+1

T flip flop on 1 toggle the input.

So on Qo = 1, Second flip flop receive 1 and toggle its output so it becomes also 1. So 11 is generated.

next Qo = 0 (again toggles its output) and 2nd flip flop recieve it and perform no change so Q1 = 1 unchanged. SO 10 generated.

In this fashion so on.

logic is on each edge trigger, 1st flip flop toggles and when 1st ff give 1 opt next ff will toggle otherwise it will remain unchanged. Very easy.

So on Qo = 1, Second flip flop receive 1 and toggle its output so it becomes also 1. So 11 is generated.

next Qo = 0 (again toggles its output) and 2nd flip flop recieve it and perform no change so Q1 = 1 unchanged. SO 10 generated.

In this fashion so on.

logic is on each edge trigger, 1st flip flop toggles and when 1st ff give 1 opt next ff will toggle otherwise it will remain unchanged. Very easy.

+5 votes

Since, Flip Flop T_{0} is clocked directly from main clock, Q_{0} will make transitions(toggle as it's T input is 1) whenever the clock pulse goes from 0 to 1 as flip flops are positive edge triggered.

Flip Flop T_{1}'s clock is depending on output of Flip Flop T_{0} which is Q_{0}. Hence, whenever Q_{0} makes a transition from 0 to 1,Q_{1}, will also make transition(toggle as it's T input is 1) .

As it it clear from timing diagram, the next four values of Q1 Q0 are 11, 10, 01 , 00. Ans (a)

+2 votes

Despite all the excellent answers explaining the in depth working of the circuit, i will like to mention a trick, that is time saving too (specific to this question only).

Here, we can see that T Flip Flop is used and T flip flop will complement the current state when presented with input one, here the Flip Flop giving the output Q0 is connected to the clock, and is given constant input 1. So, Q0 will alternate with every clock pulse. Only option A satisfies this condition.

Here, we can see that T Flip Flop is used and T flip flop will complement the current state when presented with input one, here the Flip Flop giving the output Q0 is connected to the clock, and is given constant input 1. So, Q0 will alternate with every clock pulse. Only option A satisfies this condition.

+1 vote

From the timing diagram it is clear that this counter is Mod 3 ripple down counter.

So, counting goes as 00,11,10,01,00 **(Option A)**

**Note** : Transitions only occur at the positive edge of the clock as the Flip Flops are edge triggered.

+1 vote

$Q_{0Next} = T_{0}\oplus Q_{0} = 1\oplus Q_{0} = Q_{0}'\:;\:$ For every clock

$Q_{1Next} = T_{1} \oplus Q_{1} = 1 \oplus Q_{1} = Q_{1}'\:;\: $ When $Q_{0}:0 \:to\: 1$ (Because clock is positive edge triggered)

$Q_{1}$ | $Q_{0}$ | $Q_{1Next}$ | $Q_{0Next}$ |

$0$ | $0$ | $1$ | $1$ |

$0$ | $1$ | $0$ | $0$ |

$1$ | $0$ | $0$ | $1$ |

$1$ | $1$ | $1$ | $0$ |

**State diagram:**

So, the correct answer is $(A)$.

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