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15,129 views
37 votes
37 votes

Which of the following input sequences will always generate a $1$ at the output $z$ at the end of the third cycle?

GATE2005-IT_43

  1. $\begin{array}{|l|l|}\hline \textbf{A} & \textbf{B} & \textbf{C} \\\hline  \text{0} & \text{0} & \text{0} \\\hline   \text{1} & \text{0} & \text{1} \\\hline  \text{1} & \text{1} & \text{1} \\\hline \end{array}$
  2. $\begin{array}{|l|l|}\hline \textbf{A} & \textbf{B} & \textbf{C} \\\hline  \text{1} & \text{0} & \text{1} \\\hline   \text{1} & \text{1} & \text{0} \\\hline  \text{1} & \text{1} & \text{1} \\\hline \end{array}$
  3. $\begin{array}{|l|l|}\hline \textbf{A} & \textbf{B} & \textbf{C} \\\hline  \text{0} & \text{1} & \text{1} \\\hline   \text{1} & \text{0} & \text{1} \\\hline  \text{1} & \text{1} & \text{1} \\\hline \end{array}$
  4. $\begin{array}{|l|l|}\hline \textbf{A} & \textbf{B} & \textbf{C} \\\hline  \text{0} & \text{0} & \text{1} \\\hline   \text{1} & \text{1} & \text{0} \\\hline  \text{1} & \text{1} & \text{1} \\\hline \end{array}$
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5 Answers

46 votes
46 votes
$${\begin{array}{|c|c|c|c|c|c|c|l|}\hline\\
\textbf{}&    \textbf{A}&  \textbf{B}&\bf{C}& \bf{Q_1}& \bf{Q_2} & \textbf{Z} & \textbf{Comment}\\\hline
\text{After } 1^{st} \text{Cycle} &\text{X} &\text{X} &\text{X} &\text{X} &\text{X} &\text{X} \\\hline \text{After } 2^{nd} \text{Cycle}&0&0&\text{X}&0&\text{X} & \text{X}& \text{$Q_1$ is 0 making A and B 0}\\ \hline    \text{After } 3^{rd} \text{Cycle}&\text{X} &\text{X} &1&1&1&1& \text{$Z$ is $1$ making $Q1$ and $Q2$ $1$,}\\&&&&&&&\text{Either $A$ or $B$ is $1$.} \\&&&&&&&\text{$Q1'$ of previous cycle is $1$.} \\ \hline
\end{array}}$$
The filling is done in reverse order. Here, none of the options match. So, something wrong somewhere.
edited by
4 votes
4 votes
Answer is C

Explanation:

Here given in question first O/P Z is always 1 at the end of third cycle.

(I have consider first f/f Q as Q0 and second below f/f as Q1)

So Z=1

now for Z=1

Iff Q0 and Q1 are 1.

for Q0=1:

Either A or B should 1 or Both

i.e.

A B

0 1

1 0

1 1

Now for Q1=1:

Q0'.C=1   only when Q0=0 and C=1

Now combine the Input sequences(for both Q0=1 and Q1=1):

A B C

0 1 1

1 0 1

1 1 1

 

So, C is the Answer here.
2 votes
2 votes

I am getting  Option(a) as the correct answer. If wrong please rectify it.

I have taken the initial state as 0. 

2 votes
2 votes
A B C $D_{top}$ $D_{bottom}$ $Q_{top}$ $Q_{bottom}$ Z Remark/Comment
X X X X X 1 1 1 Final Value
1 0 1 1 1 0 X X Required Value before $3^{rd}$ clock
0 0 X 0 X X X X Required Value before $2^{nd}$ clock
X X X X X X X X Required Value before $1^{st}$ clock

In the $2^{nd}$ row (from top). Following values are also possible.

0 1 1 1 1 0 X X Required Value before $3^{rd}$ clock
1 1 1 1 1 0 X X Required Value before $3^{rd}$ clock

So non of the options are matching. I hope this helps.

ping @Krish__, @rahul sharma 5,  @Red_devil, @Shivam Chauhan, @Tuhin Dutta, @Anu007,  @Ashwin Kulkarni @reena_kandari  and @srestha ji

Answer:

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