GATE IT 2005 | Question: 43

Question

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

• A. $\begin{array}{|l|l|}\hline \textbf{A} & \textbf{B} & \textbf{C} \\\hline  \text{0} & \text{0} & \text{0} \\\hline   \text{1} & \text{0} & \text{1} \\\hline  \text{1} & \text{1} & \text{1} \\\hline \end{array}$
• B. $\begin{array}{|l|l|}\hline \textbf{A} & \textbf{B} & \textbf{C} \\\hline  \text{1} & \text{0} & \text{1} \\\hline   \text{1} & \text{1} & \text{0} \\\hline  \text{1} & \text{1} & \text{1} \\\hline \end{array}$
• C. $\begin{array}{|l|l|}\hline \textbf{A} & \textbf{B} & \textbf{C} \\\hline  \text{0} & \text{1} & \text{1} \\\hline   \text{1} & \text{0} & \text{1} \\\hline  \text{1} & \text{1} & \text{1} \\\hline \end{array}$
• D. $\begin{array}{|l|l|}\hline \textbf{A} & \textbf{B} & \textbf{C} \\\hline  \text{0} & \text{0} & \text{1} \\\hline   \text{1} & \text{1} & \text{0} \\\hline  \text{1} & \text{1} & \text{1} \\\hline \end{array}$

Comments

None is correct

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If the input of the second flip-flop depends on the output of the first, and you don’t know what that first output is initially, then: You cannot determine the second flip-flop's output after the first clock cycle without checking.

Let me know if i am wrong

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@arbpass yess..option(b) and (c) get ruled out because B=1 initially and input depends on Q'..so we cannot comment anything about output. Option (a) and (c) are also wrong because z≠1 by the end of third cycle.

Answer 1

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

Comments

If you look at your explanation after 2nd clock if you give A=0 , B=0  then you will get Q1 = 0 at after 3rd cycle it doesn't make any  sense,

But if you give after 1st cycle A=0, B=0 then you will get after 2nd clock Q1=0 so the answer matched with the option A

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@Arjun @gatecse @Deepak Poonia

→ It can be concluded that A and B should be 0 for the second cycle and in the third clock cycle, either A or B or both has to be 1 and in the first clock cycle, we don't know the value of A and B.

→   After the 3rd clock cycle value of Z is 1 and after the 2nd clock cycle value of Z is 0 and after the 1st clock cycle, we don't know the value of Z.

Sir this picture is correct?

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yes I think in second row all col should be 000 or 001 and third ABC should be either 111 or 101 or 011  dosent matter what first row as Q2 depends on CQ1' so previous value of AB has to be 00 so that Q1' can produce 1 and hence after third cycle we can get 1 is that ryt?

 

Answer 2

The equation itself if the given digram says answer will always be zero no matter what A B C are, am I being sensible or something is wrong in my conclusion?

Comments

yeah, i also did the same strategy for calculating z and got the same conclusion that z cannot be 1.

Answer 3

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.

Comments

Kindly check your logical implication again..

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Yes,C is the correct answer.

Answer 4

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

I have taken the initial state as 0. 

Comments

check your x'

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after applying clock 1, x and x' can't be same bcz after applying clock after sometime they must be different according to the circuit. so in this example all option should be wrong.

Answer 5

z=${Q_{0}}$${Q_{1}}$

Option A

0 0 0 (implies) ⇒ ${Q_{0}}$ = 0   $\bar{{Q_{0}}}$=1   ${Q_{1}}$ = 0               ⇒ z=0

1 0 1 ⇒                ${Q_{0}}$ = 1   $\bar{{Q_{0}}}$=0   ${Q_{1}}$ = 1                ⇒ z=1

1 1 1 ⇒                ${Q_{0}}$ = 1   $\bar{{Q_{0}}}$=0   ${Q_{1}}$ = 0                ⇒ z=0

Hence, this is wrong.

Option B

1 0 1 ⇒                ${Q_{0}}$ = 1   $\bar{{Q_{0}}}$=0   ${Q_{1}}$ = don’t know  ⇒ z=don’t know

1 1 0 ⇒                ${Q_{0}}$ = 1   $\bar{{Q_{0}}}$=0   ${Q_{1}}$ = 1                  ⇒ z=0

1 1 1 ⇒                ${Q_{0}}$ = 1   $\bar{{Q_{0}}}$=0   ${Q_{1}}$ = 0                  ⇒ z=0

Hence, this is wrong too.

Option C

0 1 1 ⇒                ${Q_{0}}$ = 1   $\bar{{Q_{0}}}$=0   ${Q_{1}}$ = don’t know   ⇒ z=don’t know

1 0 1 ⇒                ${Q_{0}}$ = 1   $\bar{{Q_{0}}}$=0   ${Q_{1}}$ = 0                     ⇒ z=0

1 1 1 ⇒                ${Q_{0}}$ = 1   $\bar{{Q_{0}}}$=0   ${Q_{1}}$ = 0                     ⇒ z=0

Hence, this is wrong too.

Option D

0 0 1 ⇒                ${Q_{0}}$ = 0   $\bar{{Q_{0}}}$=1   ${Q_{1}}$ = don’t know   ⇒ z=don’t know

1 1 0 ⇒                ${Q_{0}}$ = 1   $\bar{{Q_{0}}}$=0   ${Q_{1}}$ = 0                     ⇒ z=0

1 1 1 ⇒                ${Q_{0}}$ = 1   $\bar{{Q_{0}}}$=0   ${Q_{1}}$ = 0                     ⇒ z=0

Hence, this is wrong too.

Answer 6

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

Comments

Whats the final conclusion on this, as 3rd cycle for all 4 options is 111 are none of the answers correct, if so was grace marks given in Gate 2005 for this questions