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The truth table shown in figure is for:
$\begin{array}{|l|l|l|l|l|} \hline A & 0 & 0 & 1 & 1 \\ \hline B & 0 & 1 & 0 & 1 \\ \hline Y & 1 & 0 & 0 & 1 \\ \hline \end{array}$
Options:
$\begin{array}{|l|l|l|l|l|} \hline A & 0 & 0 & 1 & 1 \\ \hline B & 0 & 1 & 0 & 1 \\ \hline Y & 1 & 0 & 0 & 1 \\ \hline \end{array}$
Solution:
2749 Upvotes
Verified Answer
The correct answer is:
XNOR
For 'XNOR' gate $Y=\bar{A} \bar{B}+A B$
$\begin{aligned}
& \text { i.e. } \overline{0} \cdot \overline{0}+0.0=1.1+0.0=1+0=1 \\
& \overline{0} \cdot \overline{1}+0.1=1 \cdot 0+0.1=0+0=0 \\
& \overline{1} \cdot \overline{0}+1 \cdot 0=0.1+1.0=0+0=0 \\
& \overline{1} \cdot \overline{1}+1 \cdot 1=0.0+1 \cdot 1=0+1=1
\end{aligned}$
$\begin{aligned}
& \text { i.e. } \overline{0} \cdot \overline{0}+0.0=1.1+0.0=1+0=1 \\
& \overline{0} \cdot \overline{1}+0.1=1 \cdot 0+0.1=0+0=0 \\
& \overline{1} \cdot \overline{0}+1 \cdot 0=0.1+1.0=0+0=0 \\
& \overline{1} \cdot \overline{1}+1 \cdot 1=0.0+1 \cdot 1=0+1=1
\end{aligned}$
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