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

How in option A you got z=1?

$Z = Q_{0n} . Q_{1n}$

$Z = 1 . 0=0$

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That was wrong. Thanks for pointing it out.

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None of the option is correct.

Answer 1

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 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?