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The truth-table for the two-input Ex-OR gate is (A, B are the inputs, $\mathrm{Y}$ is the output).
Options:
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2689 Upvotes
Verified Answer
The correct answer is:
S
The Boolean expression for Exclusive-OR gate is as follows:
$Y=(A \oplus B)=\bar{A} \cdot B+A \cdot \bar{B}$"ō"
The output of an Exclusive-OR gate ONLY goes "HIGH" when its two input terminals are at "DIFFERENT" logic levels with respect to each other.
Therefore, option table S is correct.
$Y=(A \oplus B)=\bar{A} \cdot B+A \cdot \bar{B}$"ō"
The output of an Exclusive-OR gate ONLY goes "HIGH" when its two input terminals are at "DIFFERENT" logic levels with respect to each other.
Therefore, option table S is correct.
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