GATE CSE 2010 | Question: 32

Question

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 $Q_1Q_0$?

 

• A. $11$, $10$, $01$, $00$
• B. $10$, $11$, $01$, $00$
• C. $10$, $00$, $01$, $11$
• D. $11$, $10$, $00$, $01$

Comments

Q0  will produce output as  0 1 0 1 0 1 ......
Q1 will produce output as   0 0 1 1 0 0 ......
 

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I think the second FF should be positive level triggered rather than positively edge triggered 

Correct me if i'm wrong !

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Answer 1

Option A.

Option A.

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

\[
\begin{array}{c|c}
T & Q_{n+1} \\
\hline
0 & Q_n \\
1 & \overline{Q_n}
\end{array}
\]

\[
\underset{\text{For rows 2 and 4, Clk for } T_1 \text{ is 0 and hence old output is retained}}{
\begin{array}{cc|cc}
T_0 & Q_0 & T_1 & Q_1 \\
\hline
  & 0 &  & 0 \\
\hline
1 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 \\
1 & 1 & 1 & 0 \\
1 & 0 & 1 & 0
\end{array}
}
\quad
\begin{array}{c|c}
Q_1 & Q_0 \\
\hline
1 & 1 \\
1 & 0 \\
0 & 1 \\
0 & 0
\end{array}
\]

Comments

here which one is q0, left or the right

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habedo007   ther is not a subtle hint ; there is a whole clock present which state that it is a flipflop 

Answer 2

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

Comments

@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 ?????

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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.

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its k @pc  :)

Answer 3

Since, Flip Flop T₀ is clocked directly from main clock, Q₀ 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 T₁'s clock is depending on output of Flip Flop T₀ which is Q₀. Hence, whenever Q₀ makes a transition from 0 to 1,Q₁, 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)

Answer 4

$Q_{0Next} = T_{0}\oplus Q_{0} = 1\oplus Q_{0} = Q_{0}'\:;\:$ 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 $(A)$.

Comments

There are so much answer but no one has given simple and clear explanation , but after solving it got your answer same as my and hence got clarity that yes I am right.

Answer 5

Asynchronous circuit, please correct me if iam wrong,.

Answer 6

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.